Question

The potential difference in volt across the resistance $R_3$ in the circuit shown In figure, is
$(R_1=15\Omega ,R_2=15\Omega ,R_3=30\Omega ,R_4=35\Omega )$

• A. $5$
• B. $7.5$
• C. $15$
• D. $2.5$

Answer 1

The correct option is C $15$

Since resistors $R_1$ and $R_2$ are connected in series.

So equivalent resistance of $R_1$ and $R_2$, $R_{1,2}=R_1+R_2=15+15=30\Omega$

Now, resistors $R_{1,2}$ and $R_3$ are connected in parallel.

So equivalent resistance of $R_{1,2}$ and $R_3$, $\frac{1}{R_{1,2,3}}=\frac{1}{R_{1,2}}+\frac{1}{R_3}=\frac{1}{30}+\frac{1}{30}$

$⟹R_{1,2,3}=15\Omega$

Since resistors $R_{1,2,3}$ and $R_4$ are connected in series.

So equivalent resistance of $R_{1,2,3}$ and $R_4$, $R_T=R_{1,2,3}+R_4=15+35=50\Omega$

Current flowing through the circuit $i=\frac{V}{R_T}=50/50=1A$

As resistances $R_{1,2}=R_3$.

Thus current flowing through each becomes $i/2$.

Current flowing through $R_3$, $i_3=\frac{1}{2}=0.5A$

Voltage across $R_3$ $V_3=i_3{R}_3=0.5\times 30=15$ volts