Question

In an electric circuit, two resistors of 12 Ω each are joined in parallel to a 6 V battery. Find the current drawn from the battery.

Answer 1

Step1: Given data

Two resistors $R_1=12\Omega ,R_2=12\Omega$ are joined in parallel

Voltage, $V=6V$

Step2: Calculating equivalent resistance in parallel

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

Substituting the value of $R_{1}and{R}_2$ in the formula

$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{12}+\frac{1}{12} \\ \frac{1}{R_{eq}}=\frac{1+1}{12} \\ \frac{1}{R_{eq}}=\frac{2}{12} \\ \frac{1}{R_{eq}}=\frac{1}{6} \\ {R}_{eq}=6\Omega \end{gathered}$$

Step3: Calculating current

$$\begin{gathered} V=I\times R[whereV=voltage,I=current,R=reis\tan ce] \\ I=\frac{V}{R}[R=R_{eq}] \\ I=\frac{6}{6} \\ I=1A \end{gathered}$$

Hence, the current drawn from the battery is $1ampere.$