Question

In the circuit given below, the voltage across R by mesh analysis is _________

(a) 1.59 V

(b) 1 V

(c) 2.54 V

(d) 1.54 V

The question was asked during an online interview.

I'm obligated to ask this question of Calculation of I(0+), di/dt(0+), d2I/dt2(0+) in Circuits Involving both Capacitor and Inductor topic in chapter Application of the Laplace Transform in Circuit Analysis of Network Theory

Answer 1

The correct choice is (d) 1.54 V

The explanation is: In loop 1, by KVL, (10 + 2) I1 + (-2) I2 + 0I3 = 5

Or, 12 I1 – 2I2 = 5

In loop 2, -2I1 + (10+2+20+2) I2 + (-2) I3 = 0

Or, -2I1 + 34 I2 – 2I3 = 0

In loop 3, 0I1 + – 2I2 + (2+10) I3 = 10

Or, 0I1 – 2I2 + 12I3 = 10

Now, the voltage across R is VR = (I2 – I3) R

Now, I2 = \(\frac{360}{4800} = \frac{3}{40}\)

Now, I3 = \(\frac{4060}{4800} = \frac{203}{240}\) A

Therefore, VR = (I2 – I3) R = \(\Big[\frac{3}{40} – \frac{203}{240}\Big] 2\) = 1.54 V.