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Let the current flowing through 5 Ω resistor be I' Applying KCL, we get
5 + I = I'.
Applying KVL, we get
So,
= 4.4 A
For the circuit shown below, match List  I with ListII and select the correct answer using the codes given below the lists:
Codes:
Nodes = 10 (a, b, c, d, e, f, g, h, k, p)
Junction points = 3(b, e, h)
The value of current I in the circuit shown below for V = 2 volt is
Applying KCL at the node connecting 0.5 V current source and 1 A current source, we get
1 + I = 0.5 V
For v  2 volt, 1 + I = 1
or, I = 0 A
Using KVL in the loop, we have
V_{1 }= 10 + 15 = 25 volts
Also, V_{2} = 25 volts
∴ V_{1} + V_{2} = 25  25 = 0 volts
The value of voltage source to be connected across the terminals X and Y so that drop across the 10 Ω resistor is 45 V is
Let the required voltage be V
Then, voltage across 10 Ω resistor
or,
Hence, required voltage is
V = 180 volts
In a voltage divider circuit as shown below with two resistances R_{1} and R_{2}, the voltage drop V_{1} is twice the drop V_{2} across R_{2}. The value of R_{1}, and R_{2} will be
Using Ohm’s law,
Given, V_{1} = 2V_{2}
As I is same therefore,
R_{1} = 2R_{2}
∴ R_{T} = R_{1} + R_{2} = 2R_{2} + R_{2} = 3R_{2}
So, R_{T} = 3R_{2} = 4Ω
or,
and
A delta connected load is shown in figure below. The equivalent star connection has a value of R in Ω is
We know that
R_{Δ }= 3 R_{Y}
or,
The voltage drop across the 2 Ω resistor for the circuit shown below is
On combining the current sources, the circuit is reduced as shown below.
Hence, voltage drop across the 2Ω resistor
= 2 x 2 = 4 volts
For the circuit shown below, the sum of the unknown currents I_{1}, I_{2}, l_{3} and I_{4} is
Using KCL,
I_{2} = 6  2 = 4 A
Also,
or,
and I_{4} = I_{2 } I_{3} = 4  (1.5) = 5.5 A
Also, I_{1} = 6 A
∴ I_{1} + I_{2} + I_{3} + I_{4} = 6 + 4  1.5 + 5.5
= 15.5  1.5 = 14A
In the delta equivalent of the given star connected circuit is equal to
= (10 + j10 + j20)
= (10 + j30) Ω
Four resistances 80 Ω, 50 Ω, 25 Ω and R are connected in parallel. Current through 25 Ω resistance is 4 A. The total current of the supply is 10 A. The value of R will be
Given, I_{25Ω}=4 A
Here, R' = 80║50║R
or,
or,
or, 29R + 400 = 40R or 11R = 400
or, R = 36.36 Ω
For the circuit shown below, the equivalent resistance will be
Here, a and c are at equipotential.
Also, b and d are at equipotential
∴
Three resistors of R Ω each are connected to form a triangle. The resistance between any two terminals will be
A network has 10 nodes and 17 branches. The number of different node pair voltage would be
Number of different nodepair voltage = Number of KCL equations = n  1 =10  1 = 9
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