Question

Two cells of 3V each are connected in parallel. An external resistance of 0.5 Ω is connected is series to the junction of two parallel resistors of 4 Ω and 2Ω and then to common terminal of battery through each resistor as shown in figure. What is the current flowing through 4Ω resistor?

A. 0.25 A B. 0.55A C. 0.35 A D. 1.50A

Answer 1

The given circuit can be reduced to: -

The equivalent resistance can be calculated as: -

R_eq = \(\frac{4.2}{4+2}\) + 0.5 = \(\frac{11}{6}\)Ω

R_eq = \(\frac{11}{6}\)Ω

The net current in the circuit is given as I = \(\frac{V}{R}\)

= \(\frac{3}{\frac{11}6}\)

= \(\frac{18}{3}\) A, so,

I_net = \(\frac{18}{3}\) A.

The circuit can be drawn as shown aside, 

The voltage drop across the parallel resistance combination is \(\frac{24}{11v}\), so net current in the 4 Ω resistor is \(\frac{6}{11v}\) = 0.55 A.