An electrical circuit is shown in Fig. The values ofresistances and the directions of the currents are shown. A voltmeter of resistance `400 Omega` is connected across the `400 Omega`. resistor. The battery has negligible internal resistance.

The value of current `i_(1)` is:

Options

A. `1/10 A`
B. `1/20 A`
C. `1/30 A`
D. `1/40 A`

Answer

C

Explanation

Resistance `R = 200 Omega` shown in the circuit is the equivalent resistance of the parallel combination of the `400 Omega` resistor and the resistance of `400 Omega` of the voltmeter.
Applying KirchhofFs loop rule to mesh I, we have
`-100 I_(2) - 100 (I_(2) - I_(3)) + 100 I_(1) =0`
or `I_(1) - 2I_(2) + I_(3)=0`........(1)
Applying Kirchhoff.s loop rule to mesh II, we have
`100 I_(1) -200(I_(1) + I_(2) + I_(3)) + 10=0`
or `3I_(1) + 2I_(2) -2I_(3) - 0.1=0`.........(2)
Applying Kirchhoff.s loop rule to mesh III, we have
`-200 I_(3) + 200(I_(1) + I_(2) - I_(3)) + 100(I_(2) - I_(3))=0`
or `2I_(1) + 3I_(2) - 5I_(3)=0`........(3)
Simultaneous solution of Eqs. (1), (2) and (3) yields.
`I_(1) = I_(2) = I_(3) = 1/30 A`