Question

In the given electric circuit if the current flowing through 3 $\Omega$ resistor is 1 A, find the voltage of the battery and the current I drawn from it.

Answer 1

Given, $R_2=3\Omega$

Current flowing through the 3-ohm resistor.

$I_2=1A$

$V_2=I_2\times R_2=1\times 3=3volts$

Resultant of $R_2$ and $R_3$ is R'

SO for the parallel combination we can write

$\frac{1}{\text{R'}}=\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{3}+\frac{1}{6}=\frac{2+1}{6}=\frac{3}{6}=\frac{1}{2}$

$R^{\prime }=2\Omega$

$V^{\prime }=V_2=3volts$

I' = ?

I' = $\frac{\text{V'}}{\text{R'}}=\frac{3}{2}$ A = 1.5 A

I' = I = 1.5 A

Resultant resistance in the circuit = R

$R=R_1+R^{\prime }=2+2=4\Omega$

I = $\frac{3}{2}$ A

So the current drawn is I = $\frac{3}{2}$ A

Now for finding voltage

V = ?

$V=IR=\frac{3}{2}\times 4=6$ volts

$\Rightarrow E=6\text{volts}$

Hence the voltage of the battery is 6 V