In the circuit shown in Figure- E-1- E-2 - E-3 - 2 - V and R-1 - R-2 - 4Omega - The current flowing between points A and B through battery E-2 iszero2 - A - from- A - to - B2 - A - from - B - to - Anone of the above

Question

In the circuit shown in Figure- E-1- E-2 - E-3 - 2 - V and R-1 - R-2 - 4Omega - The current flowing between points A and B through battery E-2  iszero2 - A - from- A - to - B2 - A - from - B - to - Anone of the above

Toppr Admin · Accepted Answer

Applying KVL -xA0-for loop -1- 2-2-4I1 or I1-1Afor loop -2- -xA0-x2212-2-x2212-2-x2212-4I2 or I2-1AThus- -I1-I2-1-1-2A will flow from A to B-xA0-

In the circuit shown in Figure, $E_{1} = E_{2} = E_{3} = 2 V$ and $R_{1} = R_{2} = 4 Ω$ . The current flowing between points $A$ and $B$ through battery $E_{2}$ is

zero
$2 A f r o m A t o B$
$2 A f r o m B t o A$
none of the above

Applying KVL ,
for loop (1), $2 + 2 = 4 I_{1}$ or $I_{1} = 1 A$
for loop (2), $- 2 - 2 = - 4 I_{2}$ or $I_{2} = 1 A$
Thus, $(I_{1} + I_{2}) = (1 + 1) = 2 A$ will flow from A to B.

In the circuit shown in Figure, E1=E2=E3=2V and R1=R2=4Ω. The current flowing between points A and B through battery E2 iszero2AfromAtoB2AfromBtoAnone of the above

Applying KVL , for loop (1), 2+2=4I1 or I1=1Afor loop (2), −2−2=−4I2 or I2=1AThus, (I1+I2)=(1+1)=2A will flow from A to B.

In the circuit shown in Figure, $E_{1} = E_{2} = E_{3} = 2 V$ and $R_{1} = R_{2} = 4 Ω$ . The current flowing between points $A$ and $B$ through battery $E_{2}$ is

zero
$2 A f r o m A t o B$
$2 A f r o m B t o A$
none of the above

Applying KVL ,
for loop (1), $2 + 2 = 4 I_{1}$ or $I_{1} = 1 A$
for loop (2), $- 2 - 2 = - 4 I_{2}$ or $I_{2} = 1 A$
Thus, $(I_{1} + I_{2}) = (1 + 1) = 2 A$ will flow from A to B.