# 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

24.9k views
in Physics
closed

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

+1 vote
by (51.1k points)
selected by

The given circuit can be reduced to: -

The equivalent resistance can be calculated as: -

Req $\frac{4.2}{4+2}$ + 0.5 = $\frac{11}{6}$Ω

Req $\frac{11}{6}$Ω

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

$\frac{3}{\frac{11}6}$

$\frac{18}{3}$ A, so,

Inet $\frac{18}{3}$ 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.