JEE Main Previous Year Questions (2025): Current Electricity

 Page 1


JEE Main Previous Year Questions 
(2025): Current Electricity 
Q1: The net current flowing in the given circuit is ____ A. 
 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 1 
Solution: 
 
R
eq
= 2O
I =
2
2
= 1 A
 
Q2: A wire of resistance ???? is bent to form an equilateral triangle. Then the 
equivalent resistance across any two vertices will be ____ ohm. 
JEE Main 2025 (Online) 24th January Morning Shift 
Ans: 2 
Solution: 
Each side of the triangle has a resistance of ?? side 
=
9O
3
= 3O . 
When measuring the equivalent resistance between any two vertices, there are two paths: 
Page 2


JEE Main Previous Year Questions 
(2025): Current Electricity 
Q1: The net current flowing in the given circuit is ____ A. 
 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 1 
Solution: 
 
R
eq
= 2O
I =
2
2
= 1 A
 
Q2: A wire of resistance ???? is bent to form an equilateral triangle. Then the 
equivalent resistance across any two vertices will be ____ ohm. 
JEE Main 2025 (Online) 24th January Morning Shift 
Ans: 2 
Solution: 
Each side of the triangle has a resistance of ?? side 
=
9O
3
= 3O . 
When measuring the equivalent resistance between any two vertices, there are two paths: 
A direct path along one side with resistance: 
?? 1
= 3O . 
An indirect path passing through the other two sides in series: 
?? 2
= 3O+ 3O = 6O. 
These two paths are in parallel, so the equivalent resistance is calculated by: 
1
?? eq 
=
1
?? 1
+
1
?? 2
=
1
3O
+
1
6O
=
2
6O
+
1
6O
=
3
6O
. 
Thus, 
?? eq 
=
6O
3
= 2O . 
The equivalent resistance across any two vertices of the equilateral triangle is therefore: 
2O . 
Q3: The value of current ?? in the electrical circuit as given below, when potential at ?? 
is equal to the potential at ?? , will be ____ A. 
 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 2 
Solution: 
 
Page 3


JEE Main Previous Year Questions 
(2025): Current Electricity 
Q1: The net current flowing in the given circuit is ____ A. 
 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 1 
Solution: 
 
R
eq
= 2O
I =
2
2
= 1 A
 
Q2: A wire of resistance ???? is bent to form an equilateral triangle. Then the 
equivalent resistance across any two vertices will be ____ ohm. 
JEE Main 2025 (Online) 24th January Morning Shift 
Ans: 2 
Solution: 
Each side of the triangle has a resistance of ?? side 
=
9O
3
= 3O . 
When measuring the equivalent resistance between any two vertices, there are two paths: 
A direct path along one side with resistance: 
?? 1
= 3O . 
An indirect path passing through the other two sides in series: 
?? 2
= 3O+ 3O = 6O. 
These two paths are in parallel, so the equivalent resistance is calculated by: 
1
?? eq 
=
1
?? 1
+
1
?? 2
=
1
3O
+
1
6O
=
2
6O
+
1
6O
=
3
6O
. 
Thus, 
?? eq 
=
6O
3
= 2O . 
The equivalent resistance across any two vertices of the equilateral triangle is therefore: 
2O . 
Q3: The value of current ?? in the electrical circuit as given below, when potential at ?? 
is equal to the potential at ?? , will be ____ A. 
 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 2 
Solution: 
 
V
A
= V
B
? the bridge is balanced 
 ?
10
R
=
20
40
R = 20O
I =
40
20
= 2 A
 
Q4: In the figure shown below, a resistance of ?????? . ???? is connected in series to an 
ammeter ?? of resistance ???????? . A shunt resistance of ?????? is connected in parallel 
with the ammeter. The reading of the ammeter is ____ mA . 
 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 5 
Solution: 
 
Page 4


JEE Main Previous Year Questions 
(2025): Current Electricity 
Q1: The net current flowing in the given circuit is ____ A. 
 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 1 
Solution: 
 
R
eq
= 2O
I =
2
2
= 1 A
 
Q2: A wire of resistance ???? is bent to form an equilateral triangle. Then the 
equivalent resistance across any two vertices will be ____ ohm. 
JEE Main 2025 (Online) 24th January Morning Shift 
Ans: 2 
Solution: 
Each side of the triangle has a resistance of ?? side 
=
9O
3
= 3O . 
When measuring the equivalent resistance between any two vertices, there are two paths: 
A direct path along one side with resistance: 
?? 1
= 3O . 
An indirect path passing through the other two sides in series: 
?? 2
= 3O+ 3O = 6O. 
These two paths are in parallel, so the equivalent resistance is calculated by: 
1
?? eq 
=
1
?? 1
+
1
?? 2
=
1
3O
+
1
6O
=
2
6O
+
1
6O
=
3
6O
. 
Thus, 
?? eq 
=
6O
3
= 2O . 
The equivalent resistance across any two vertices of the equilateral triangle is therefore: 
2O . 
Q3: The value of current ?? in the electrical circuit as given below, when potential at ?? 
is equal to the potential at ?? , will be ____ A. 
 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 2 
Solution: 
 
V
A
= V
B
? the bridge is balanced 
 ?
10
R
=
20
40
R = 20O
I =
40
20
= 2 A
 
Q4: In the figure shown below, a resistance of ?????? . ???? is connected in series to an 
ammeter ?? of resistance ???????? . A shunt resistance of ?????? is connected in parallel 
with the ammeter. The reading of the ammeter is ____ mA . 
 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 5 
Solution: 
 
R
eq
= R
1
+ R
2
R
eq
= 150.4 +
240 × 10
250
 = 150.4 + 9.6 = 160O
I
1
=
IR
2
240
I
1
=
I × 9.6
240
 =
20
160
×
9.6
2400
=
1
200
= 5 × 10
-3
 A = 5 mA
 
Q5: Two cells of emfs 1 V and 2 V and internal resistances ???? and ???? , respectively, 
are connected in series with an external resistance of ???? . The total current in the 
circuit is ?? ?? . Now the same two cells in parallel configuration are connected to same 
external resistance. In this case, the total current drawn is ?? ?? . The value of (
?? ?? ?? ?? ) is 
?? ?? . 
The value of ?? is ____ . 
JEE Main 2025 (Online) 3rd April Evening Shift 
Ans: 4 
Solution: 
 
?? cq
= 3
R
cq
= 9
i
1
=
3
9
=
1
3
 
Page 5


JEE Main Previous Year Questions 
(2025): Current Electricity 
Q1: The net current flowing in the given circuit is ____ A. 
 
JEE Main 2025 (Online) 22nd January Evening Shift 
Ans: 1 
Solution: 
 
R
eq
= 2O
I =
2
2
= 1 A
 
Q2: A wire of resistance ???? is bent to form an equilateral triangle. Then the 
equivalent resistance across any two vertices will be ____ ohm. 
JEE Main 2025 (Online) 24th January Morning Shift 
Ans: 2 
Solution: 
Each side of the triangle has a resistance of ?? side 
=
9O
3
= 3O . 
When measuring the equivalent resistance between any two vertices, there are two paths: 
A direct path along one side with resistance: 
?? 1
= 3O . 
An indirect path passing through the other two sides in series: 
?? 2
= 3O+ 3O = 6O. 
These two paths are in parallel, so the equivalent resistance is calculated by: 
1
?? eq 
=
1
?? 1
+
1
?? 2
=
1
3O
+
1
6O
=
2
6O
+
1
6O
=
3
6O
. 
Thus, 
?? eq 
=
6O
3
= 2O . 
The equivalent resistance across any two vertices of the equilateral triangle is therefore: 
2O . 
Q3: The value of current ?? in the electrical circuit as given below, when potential at ?? 
is equal to the potential at ?? , will be ____ A. 
 
JEE Main 2025 (Online) 28th January Evening Shift 
Ans: 2 
Solution: 
 
V
A
= V
B
? the bridge is balanced 
 ?
10
R
=
20
40
R = 20O
I =
40
20
= 2 A
 
Q4: In the figure shown below, a resistance of ?????? . ???? is connected in series to an 
ammeter ?? of resistance ???????? . A shunt resistance of ?????? is connected in parallel 
with the ammeter. The reading of the ammeter is ____ mA . 
 
JEE Main 2025 (Online) 3rd April Morning Shift 
Ans: 5 
Solution: 
 
R
eq
= R
1
+ R
2
R
eq
= 150.4 +
240 × 10
250
 = 150.4 + 9.6 = 160O
I
1
=
IR
2
240
I
1
=
I × 9.6
240
 =
20
160
×
9.6
2400
=
1
200
= 5 × 10
-3
 A = 5 mA
 
Q5: Two cells of emfs 1 V and 2 V and internal resistances ???? and ???? , respectively, 
are connected in series with an external resistance of ???? . The total current in the 
circuit is ?? ?? . Now the same two cells in parallel configuration are connected to same 
external resistance. In this case, the total current drawn is ?? ?? . The value of (
?? ?? ?? ?? ) is 
?? ?? . 
The value of ?? is ____ . 
JEE Main 2025 (Online) 3rd April Evening Shift 
Ans: 4 
Solution: 
 
?? cq
= 3
R
cq
= 9
i
1
=
3
9
=
1
3
 
 
?? eq 
=
?? 1
r
1
+
?? 2
r
2
1
r
1
+
1
r
2
?? eq 
=
1
2
+
2
1
1
2
+
1
1
=
5
3
r
equ 
=
2 × 1
3
+ 6 =
20
3
i
2
=
1
4
?
i
1
i
2
=
4
3
 
Q6:  
 
Sliding contact of a potentiometer is in the middle of the potentiometer wire having 
resistance ?? ?? = ???? as shown in the figure. An external resistance of ?? ?? = ???? is 
connected via the sliding contact. The electric current in the circuit is :