JEE Main Previous Year Questions (2025): Alternating Currents

 Page 1


JEE Main Previous Year Questions 
(2025): Alternating Currents 
 
Q1:A series LCR circuit is connected to an alternating source of emf ?? . The current 
amplitude at resonant frequency is ?? ?? . If the value of resistance ?? becomes twice of its 
initial value then amplitude of current at resonance will be 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 
I
0
2
 
B. 
I
0
v 2
 
C. 2 I
0
 
D. I
0
 
Ans: A 
Solution: 
At resonance in a series LCR circuit, the reactive components (inductive and capacitive) cancel 
each other out, leaving only the resistance. Therefore, the amplitude of the current is given by: 
?? 0
=
?? ?? 
Now, if the resistance is doubled, the new resistance becomes 2 ?? . The current amplitude at 
resonance then becomes: 
?? =
?? 2 ?? =
1
2
(
?? ?? ) =
?? 0
2
 
Thus, the amplitude of current at resonance when the resistance is doubled is 
?? 0
2
. 
The correct option is A. 
Q2: An alternating current is given by ?? = ?? ?? ?? ???? ? ???? + ?? ?? ?? ?? ?? ? ???? . The r.m.s current will 
be 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 
v ?? ?? 2
+ ?? ?? 2
2
 
B. 
v
I
A
2
+ I
B
2
2
 
C. 
| I
A
+ I
B
|
v 2
 
D. v I
A
2
+ I
B
2
 
Page 2


JEE Main Previous Year Questions 
(2025): Alternating Currents 
 
Q1:A series LCR circuit is connected to an alternating source of emf ?? . The current 
amplitude at resonant frequency is ?? ?? . If the value of resistance ?? becomes twice of its 
initial value then amplitude of current at resonance will be 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 
I
0
2
 
B. 
I
0
v 2
 
C. 2 I
0
 
D. I
0
 
Ans: A 
Solution: 
At resonance in a series LCR circuit, the reactive components (inductive and capacitive) cancel 
each other out, leaving only the resistance. Therefore, the amplitude of the current is given by: 
?? 0
=
?? ?? 
Now, if the resistance is doubled, the new resistance becomes 2 ?? . The current amplitude at 
resonance then becomes: 
?? =
?? 2 ?? =
1
2
(
?? ?? ) =
?? 0
2
 
Thus, the amplitude of current at resonance when the resistance is doubled is 
?? 0
2
. 
The correct option is A. 
Q2: An alternating current is given by ?? = ?? ?? ?? ???? ? ???? + ?? ?? ?? ?? ?? ? ???? . The r.m.s current will 
be 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 
v ?? ?? 2
+ ?? ?? 2
2
 
B. 
v
I
A
2
+ I
B
2
2
 
C. 
| I
A
+ I
B
|
v 2
 
D. v I
A
2
+ I
B
2
 
Ans: B 
Solution: 
?? r ms
=
v
?? ?? 2
+ ?? ?? 2
2
 
To determine the root mean square (r.m.s) value of the alternating current given by 
?? ( ?? ) = ?? ?? s i n ? ( ???? ) + ?? ?? c os ? ( ???? ) 
we start by squaring the current: 
?? ( ?? )
2
= ?? ?? 2
s i n
2
? ( ???? ) + ?? ?? 2
c os
2
? ( ???? ) + 2 ?? ?? ?? ?? s i n ? ( ???? ) c os ? ( ???? ) 
The r.m.s value is defined as the square root of the average of this squared current over one 
complete period. When averaging over a full cycle, we utilize the known averages: 
? sin
2
? ( ???? ) ? =
1
2
, ? c os
2
? ( ???? ) ? =
1
2
, ?sin ? ( ???? ) c os ? ( ???? ) ? = 0 . 
Thus, the time-averaged square of the current becomes: 
? ?? ( ?? )
2
? = ?? ?? 2
1
2
+ ?? ?? 2
1
2
=
?? ?? 2
+ ?? ?? 2
2
 
Taking the square root gives the r.m.s current: 
?? r ms
=
v
?? ?
2
+ ?? ?? 2
2
 
This corresponds to Option B. 
Q3: An electric bulb rated as ?????? ?? - ?????? ?? is connected to an ac source of rms 
voltage 220 V . The peak value of current through the bulb is : 
JEE Main 2025 (Online) 3rd April Evening Shift 
Options: 
A. 0.32 A 
B. 0.64 A 
C. 0.45 A 
D. 2.2 A 
Ans: B 
Solution: 
First find the bulb's rms current and then convert to its peak value: 
rms current 
Page 3


JEE Main Previous Year Questions 
(2025): Alternating Currents 
 
Q1:A series LCR circuit is connected to an alternating source of emf ?? . The current 
amplitude at resonant frequency is ?? ?? . If the value of resistance ?? becomes twice of its 
initial value then amplitude of current at resonance will be 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 
I
0
2
 
B. 
I
0
v 2
 
C. 2 I
0
 
D. I
0
 
Ans: A 
Solution: 
At resonance in a series LCR circuit, the reactive components (inductive and capacitive) cancel 
each other out, leaving only the resistance. Therefore, the amplitude of the current is given by: 
?? 0
=
?? ?? 
Now, if the resistance is doubled, the new resistance becomes 2 ?? . The current amplitude at 
resonance then becomes: 
?? =
?? 2 ?? =
1
2
(
?? ?? ) =
?? 0
2
 
Thus, the amplitude of current at resonance when the resistance is doubled is 
?? 0
2
. 
The correct option is A. 
Q2: An alternating current is given by ?? = ?? ?? ?? ???? ? ???? + ?? ?? ?? ?? ?? ? ???? . The r.m.s current will 
be 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 
v ?? ?? 2
+ ?? ?? 2
2
 
B. 
v
I
A
2
+ I
B
2
2
 
C. 
| I
A
+ I
B
|
v 2
 
D. v I
A
2
+ I
B
2
 
Ans: B 
Solution: 
?? r ms
=
v
?? ?? 2
+ ?? ?? 2
2
 
To determine the root mean square (r.m.s) value of the alternating current given by 
?? ( ?? ) = ?? ?? s i n ? ( ???? ) + ?? ?? c os ? ( ???? ) 
we start by squaring the current: 
?? ( ?? )
2
= ?? ?? 2
s i n
2
? ( ???? ) + ?? ?? 2
c os
2
? ( ???? ) + 2 ?? ?? ?? ?? s i n ? ( ???? ) c os ? ( ???? ) 
The r.m.s value is defined as the square root of the average of this squared current over one 
complete period. When averaging over a full cycle, we utilize the known averages: 
? sin
2
? ( ???? ) ? =
1
2
, ? c os
2
? ( ???? ) ? =
1
2
, ?sin ? ( ???? ) c os ? ( ???? ) ? = 0 . 
Thus, the time-averaged square of the current becomes: 
? ?? ( ?? )
2
? = ?? ?? 2
1
2
+ ?? ?? 2
1
2
=
?? ?? 2
+ ?? ?? 2
2
 
Taking the square root gives the r.m.s current: 
?? r ms
=
v
?? ?
2
+ ?? ?? 2
2
 
This corresponds to Option B. 
Q3: An electric bulb rated as ?????? ?? - ?????? ?? is connected to an ac source of rms 
voltage 220 V . The peak value of current through the bulb is : 
JEE Main 2025 (Online) 3rd April Evening Shift 
Options: 
A. 0.32 A 
B. 0.64 A 
C. 0.45 A 
D. 2.2 A 
Ans: B 
Solution: 
First find the bulb's rms current and then convert to its peak value: 
rms current 
?? r ms
=
?? ?? =
100 W
220 V
˜ 0 . 455 A 
peak current 
?? peak 
= ?? rms 
v 2 ˜ 0 . 455 × 1 . 414 ˜ 0 . 64 A 
So the correct choice is ?? . ?????? (Option B). 
Q4: An alternating current is represented by the equation, ?? = ?????? v ?? ?? ???? ? ( ?????? ???? ) 
ampere. The RMS value of current and the frequency of the given alternating current 
are 
JEE Main 2025 (Online) 4th April Morning Shift 
Options: 
A. 
100
v 2
 A , 100 Hz 
B. 50 v 2 A , 50 Hz 
C. 100 v 2 A , 100 Hz 
D. 100 A , 50 Hz 
Ans: D 
Solution: 
To solve the problem, we need to determine both the RMS value of the current and the 
frequency from the given equation: 
?? ( ?? ) = 100 v 2 s i n ? ( 100 ???? ) . 
Here's how we do it step by step: 
RMS Current Calculation: 
For a sinusoidal current of the form ?? ( ?? ) = ?? 0
s i n ? ( ???? ) , the RMS (root-mean-square) current is 
given by: 
?? RMS 
=
?? 0
v 2
. 
In our case, the amplitude ?? 0
= 100 v 2 A . Plugging into the formula: 
?? RMS 
=
100 v 2
v 2
= 100 A . 
Frequency Calculation: 
The argument of the sine function is 100 ???? . In the general form sin ? ( ???? ) , ?? is the angular 
frequency which relates to the frequency ?? by the equation: 
?? = 2 ???? . 
Given ?? = 100 ?? , we can solve for ?? : 
?? =
?? 2 ?? =
100 ?? 2 ?? = 50 Hz. 
With these calculations, we find that the RMS current is 100 A and the frequency is 50 Hz . 
Thus, the correct option is: 
Option D: 100 A , 50 Hz. 
Page 4


JEE Main Previous Year Questions 
(2025): Alternating Currents 
 
Q1:A series LCR circuit is connected to an alternating source of emf ?? . The current 
amplitude at resonant frequency is ?? ?? . If the value of resistance ?? becomes twice of its 
initial value then amplitude of current at resonance will be 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 
I
0
2
 
B. 
I
0
v 2
 
C. 2 I
0
 
D. I
0
 
Ans: A 
Solution: 
At resonance in a series LCR circuit, the reactive components (inductive and capacitive) cancel 
each other out, leaving only the resistance. Therefore, the amplitude of the current is given by: 
?? 0
=
?? ?? 
Now, if the resistance is doubled, the new resistance becomes 2 ?? . The current amplitude at 
resonance then becomes: 
?? =
?? 2 ?? =
1
2
(
?? ?? ) =
?? 0
2
 
Thus, the amplitude of current at resonance when the resistance is doubled is 
?? 0
2
. 
The correct option is A. 
Q2: An alternating current is given by ?? = ?? ?? ?? ???? ? ???? + ?? ?? ?? ?? ?? ? ???? . The r.m.s current will 
be 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 
v ?? ?? 2
+ ?? ?? 2
2
 
B. 
v
I
A
2
+ I
B
2
2
 
C. 
| I
A
+ I
B
|
v 2
 
D. v I
A
2
+ I
B
2
 
Ans: B 
Solution: 
?? r ms
=
v
?? ?? 2
+ ?? ?? 2
2
 
To determine the root mean square (r.m.s) value of the alternating current given by 
?? ( ?? ) = ?? ?? s i n ? ( ???? ) + ?? ?? c os ? ( ???? ) 
we start by squaring the current: 
?? ( ?? )
2
= ?? ?? 2
s i n
2
? ( ???? ) + ?? ?? 2
c os
2
? ( ???? ) + 2 ?? ?? ?? ?? s i n ? ( ???? ) c os ? ( ???? ) 
The r.m.s value is defined as the square root of the average of this squared current over one 
complete period. When averaging over a full cycle, we utilize the known averages: 
? sin
2
? ( ???? ) ? =
1
2
, ? c os
2
? ( ???? ) ? =
1
2
, ?sin ? ( ???? ) c os ? ( ???? ) ? = 0 . 
Thus, the time-averaged square of the current becomes: 
? ?? ( ?? )
2
? = ?? ?? 2
1
2
+ ?? ?? 2
1
2
=
?? ?? 2
+ ?? ?? 2
2
 
Taking the square root gives the r.m.s current: 
?? r ms
=
v
?? ?
2
+ ?? ?? 2
2
 
This corresponds to Option B. 
Q3: An electric bulb rated as ?????? ?? - ?????? ?? is connected to an ac source of rms 
voltage 220 V . The peak value of current through the bulb is : 
JEE Main 2025 (Online) 3rd April Evening Shift 
Options: 
A. 0.32 A 
B. 0.64 A 
C. 0.45 A 
D. 2.2 A 
Ans: B 
Solution: 
First find the bulb's rms current and then convert to its peak value: 
rms current 
?? r ms
=
?? ?? =
100 W
220 V
˜ 0 . 455 A 
peak current 
?? peak 
= ?? rms 
v 2 ˜ 0 . 455 × 1 . 414 ˜ 0 . 64 A 
So the correct choice is ?? . ?????? (Option B). 
Q4: An alternating current is represented by the equation, ?? = ?????? v ?? ?? ???? ? ( ?????? ???? ) 
ampere. The RMS value of current and the frequency of the given alternating current 
are 
JEE Main 2025 (Online) 4th April Morning Shift 
Options: 
A. 
100
v 2
 A , 100 Hz 
B. 50 v 2 A , 50 Hz 
C. 100 v 2 A , 100 Hz 
D. 100 A , 50 Hz 
Ans: D 
Solution: 
To solve the problem, we need to determine both the RMS value of the current and the 
frequency from the given equation: 
?? ( ?? ) = 100 v 2 s i n ? ( 100 ???? ) . 
Here's how we do it step by step: 
RMS Current Calculation: 
For a sinusoidal current of the form ?? ( ?? ) = ?? 0
s i n ? ( ???? ) , the RMS (root-mean-square) current is 
given by: 
?? RMS 
=
?? 0
v 2
. 
In our case, the amplitude ?? 0
= 100 v 2 A . Plugging into the formula: 
?? RMS 
=
100 v 2
v 2
= 100 A . 
Frequency Calculation: 
The argument of the sine function is 100 ???? . In the general form sin ? ( ???? ) , ?? is the angular 
frequency which relates to the frequency ?? by the equation: 
?? = 2 ???? . 
Given ?? = 100 ?? , we can solve for ?? : 
?? =
?? 2 ?? =
100 ?? 2 ?? = 50 Hz. 
With these calculations, we find that the RMS current is 100 A and the frequency is 50 Hz . 
Thus, the correct option is: 
Option D: 100 A , 50 Hz. 
Q5: An ac current is represented as 
?? = ?? v ?? + ???? ?? ?? ?? ? ( ?????? ???? +
?? ?? ) ?????? 
The r.m.s value of the current is 
JEE Main 2025 (Online) 7th April Morning Shift 
Options: 
A. 10 Amp 
B. 5 v 2 A m p 
C. 100 Amp 
D. 50 Amp 
Ans: A 
Solution: 
To find the root mean square (RMS) value of the given alternating current, follow these steps: 
The current is represented as: 
?? = 5 v 2 + 10c os ? ( 650 ???? +
?? 6
) A m p 
Here, the time-independent DC component is 5 v 2 and the AC component is 10c o s ? ( 650 ???? +
?? 6
). 
Calculate the square of the current, ?? 2
 : 
?? 2
= ( 5 v 2 )
2
+ ( 10c os ? ( 650 ???? +
?? 6
) )
2
+ 2 × 5 v 2 × 10c os ? ( 650 ???? +
?? 6
) 
Simplifying, we have: 
?? 2
= 50 + 100 c os
2
? ( 650 ???? +
?? 6
) + 100 v 2 c o s ? ( 650 ???? +
?? 6
) 
Find the average value ? ?? 2
? : 
The average value of c os terms over a period is zero, simplifying our equation to: 
? ?? 2
? = 50 +
100
2
+ 0 
This simplifies to: 
? ?? 2
? = 50 + 50 = 100 
Calculate the RMS current: 
The RMS value is the square root of the mean of the squares of the current: 
? ?? ? = v 100 = 10Amp 
Thus, the RMS value of the current is 10 Amps . 
Q6: In a series LCR circuit, a resistor of ???????? , a capacitor of 25 nF and an inductor of 
?????? ???? are used. For maximum current in the circuit, the angular frequency of the ac 
source is _ _ _ _ × ????
?? radians ?? - ?? 
Page 5


JEE Main Previous Year Questions 
(2025): Alternating Currents 
 
Q1:A series LCR circuit is connected to an alternating source of emf ?? . The current 
amplitude at resonant frequency is ?? ?? . If the value of resistance ?? becomes twice of its 
initial value then amplitude of current at resonance will be 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. 
I
0
2
 
B. 
I
0
v 2
 
C. 2 I
0
 
D. I
0
 
Ans: A 
Solution: 
At resonance in a series LCR circuit, the reactive components (inductive and capacitive) cancel 
each other out, leaving only the resistance. Therefore, the amplitude of the current is given by: 
?? 0
=
?? ?? 
Now, if the resistance is doubled, the new resistance becomes 2 ?? . The current amplitude at 
resonance then becomes: 
?? =
?? 2 ?? =
1
2
(
?? ?? ) =
?? 0
2
 
Thus, the amplitude of current at resonance when the resistance is doubled is 
?? 0
2
. 
The correct option is A. 
Q2: An alternating current is given by ?? = ?? ?? ?? ???? ? ???? + ?? ?? ?? ?? ?? ? ???? . The r.m.s current will 
be 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. 
v ?? ?? 2
+ ?? ?? 2
2
 
B. 
v
I
A
2
+ I
B
2
2
 
C. 
| I
A
+ I
B
|
v 2
 
D. v I
A
2
+ I
B
2
 
Ans: B 
Solution: 
?? r ms
=
v
?? ?? 2
+ ?? ?? 2
2
 
To determine the root mean square (r.m.s) value of the alternating current given by 
?? ( ?? ) = ?? ?? s i n ? ( ???? ) + ?? ?? c os ? ( ???? ) 
we start by squaring the current: 
?? ( ?? )
2
= ?? ?? 2
s i n
2
? ( ???? ) + ?? ?? 2
c os
2
? ( ???? ) + 2 ?? ?? ?? ?? s i n ? ( ???? ) c os ? ( ???? ) 
The r.m.s value is defined as the square root of the average of this squared current over one 
complete period. When averaging over a full cycle, we utilize the known averages: 
? sin
2
? ( ???? ) ? =
1
2
, ? c os
2
? ( ???? ) ? =
1
2
, ?sin ? ( ???? ) c os ? ( ???? ) ? = 0 . 
Thus, the time-averaged square of the current becomes: 
? ?? ( ?? )
2
? = ?? ?? 2
1
2
+ ?? ?? 2
1
2
=
?? ?? 2
+ ?? ?? 2
2
 
Taking the square root gives the r.m.s current: 
?? r ms
=
v
?? ?
2
+ ?? ?? 2
2
 
This corresponds to Option B. 
Q3: An electric bulb rated as ?????? ?? - ?????? ?? is connected to an ac source of rms 
voltage 220 V . The peak value of current through the bulb is : 
JEE Main 2025 (Online) 3rd April Evening Shift 
Options: 
A. 0.32 A 
B. 0.64 A 
C. 0.45 A 
D. 2.2 A 
Ans: B 
Solution: 
First find the bulb's rms current and then convert to its peak value: 
rms current 
?? r ms
=
?? ?? =
100 W
220 V
˜ 0 . 455 A 
peak current 
?? peak 
= ?? rms 
v 2 ˜ 0 . 455 × 1 . 414 ˜ 0 . 64 A 
So the correct choice is ?? . ?????? (Option B). 
Q4: An alternating current is represented by the equation, ?? = ?????? v ?? ?? ???? ? ( ?????? ???? ) 
ampere. The RMS value of current and the frequency of the given alternating current 
are 
JEE Main 2025 (Online) 4th April Morning Shift 
Options: 
A. 
100
v 2
 A , 100 Hz 
B. 50 v 2 A , 50 Hz 
C. 100 v 2 A , 100 Hz 
D. 100 A , 50 Hz 
Ans: D 
Solution: 
To solve the problem, we need to determine both the RMS value of the current and the 
frequency from the given equation: 
?? ( ?? ) = 100 v 2 s i n ? ( 100 ???? ) . 
Here's how we do it step by step: 
RMS Current Calculation: 
For a sinusoidal current of the form ?? ( ?? ) = ?? 0
s i n ? ( ???? ) , the RMS (root-mean-square) current is 
given by: 
?? RMS 
=
?? 0
v 2
. 
In our case, the amplitude ?? 0
= 100 v 2 A . Plugging into the formula: 
?? RMS 
=
100 v 2
v 2
= 100 A . 
Frequency Calculation: 
The argument of the sine function is 100 ???? . In the general form sin ? ( ???? ) , ?? is the angular 
frequency which relates to the frequency ?? by the equation: 
?? = 2 ???? . 
Given ?? = 100 ?? , we can solve for ?? : 
?? =
?? 2 ?? =
100 ?? 2 ?? = 50 Hz. 
With these calculations, we find that the RMS current is 100 A and the frequency is 50 Hz . 
Thus, the correct option is: 
Option D: 100 A , 50 Hz. 
Q5: An ac current is represented as 
?? = ?? v ?? + ???? ?? ?? ?? ? ( ?????? ???? +
?? ?? ) ?????? 
The r.m.s value of the current is 
JEE Main 2025 (Online) 7th April Morning Shift 
Options: 
A. 10 Amp 
B. 5 v 2 A m p 
C. 100 Amp 
D. 50 Amp 
Ans: A 
Solution: 
To find the root mean square (RMS) value of the given alternating current, follow these steps: 
The current is represented as: 
?? = 5 v 2 + 10c os ? ( 650 ???? +
?? 6
) A m p 
Here, the time-independent DC component is 5 v 2 and the AC component is 10c o s ? ( 650 ???? +
?? 6
). 
Calculate the square of the current, ?? 2
 : 
?? 2
= ( 5 v 2 )
2
+ ( 10c os ? ( 650 ???? +
?? 6
) )
2
+ 2 × 5 v 2 × 10c os ? ( 650 ???? +
?? 6
) 
Simplifying, we have: 
?? 2
= 50 + 100 c os
2
? ( 650 ???? +
?? 6
) + 100 v 2 c o s ? ( 650 ???? +
?? 6
) 
Find the average value ? ?? 2
? : 
The average value of c os terms over a period is zero, simplifying our equation to: 
? ?? 2
? = 50 +
100
2
+ 0 
This simplifies to: 
? ?? 2
? = 50 + 50 = 100 
Calculate the RMS current: 
The RMS value is the square root of the mean of the squares of the current: 
? ?? ? = v 100 = 10Amp 
Thus, the RMS value of the current is 10 Amps . 
Q6: In a series LCR circuit, a resistor of ???????? , a capacitor of 25 nF and an inductor of 
?????? ???? are used. For maximum current in the circuit, the angular frequency of the ac 
source is _ _ _ _ × ????
?? radians ?? - ?? 
JEE Main 2025 (Online) 23rd January Evening Shift 
Ans: 2 
Solution: 
?? =
1
v ????
 
For a series LCR circuit, the maximum current occurs at resonance, where the inductive 
reactance equals the capacitive reactance. Given the values: 
Inductance: ?? = 100m H = 0 . 1H 
Capacitance: ?? = 25nF = 25 × 10
- 9
 F 
we first calculate the product ???? : 
???? = 0 . 1 × 25 × 10
- 9
= 2 . 5 × 10
- 9
 
Next, compute the square root of the product: 
v ???? =
v
2 . 5 × 10
- 9
 
Recognize that: 
v
2 . 5 × 10
- 9
= v 2 . 5 ×
v
10
- 9
˜ 1 . 581 × 10
- 4 . 5
 
Since 10
- 4 . 5
= 3 . 162 × 10
- 5
, we have: 
v ???? ˜ 1 . 581 × 3 . 162 × 10
- 5
˜ 5 . 0 × 10
- 5
 
Now, the angular frequency at resonance becomes: 
?? =
1
5 . 0 × 10
- 5
= 2 . 0 × 10
4
 radians / s 
Thus, the angular frequency of the AC source for maximum current in the circuit is: 
?? = 2 × 10
4
 radians/s 
Q7: An inductor of self inductance 1 H is connected in series with a resistor of 
?????? ?? ?? ???? and an ac supply of ?????? ?? volt, 50 Hz . Maximum current flowing in the 
circuit is _ _ _ _ A. 
JEE Main 2025 (Online) 4th April Evening Shift 
Ans: 1 
Solution: 
Impedance of circuit 
?? ? = v ?? 2
+ ( ?? ?? )
2
= v ?? 2
+ ( ?? ?? )
2
? = v ( 100 ?? )
2
+ ( 2 ?? × 50 × 1 )
2
? = v ( 100 ?? )
2
+ ( 100 ?? )
2
? = v 2 × 100 ?? ?? ?? ?? ?? ? =
?? 2
=
100 ?? v 2 × 100 ?? =
1
v 2
?? max
? = v 2 ?? ?? ?? ?? = v 2 ×
1
v 2
= 1 Ampere 
 
Correct Ans: 1