In the circuit shown in Figure, mark the correct options.

Potential drop across

R_{1}

3.2 V .

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Potential drop across

R_{2}

5.4 V .

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Potential drop across

R_{1}

7.2 V .

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Potential drop across

R_{2}

4.8 V .

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Solution

The correct options are
C Potential drop across $R_{1}$ is $7.2 V .$
D Potential drop across $R_{2}$ is $4.8 V .$
Here the points a,h,g and f have the same potential. They are connected by conducting wires without any circuit elements between them. Similarly, points b,c,d, and e have the same potential. Hence, the potential drop across branch e and f, and a and b is the same. The two resistors ( $R_{1} = 6 Ω$ and $R_{2} = 4 Ω$ in series) are directly connected across the terminals of 12 V battery.
Thus the current flow in resistors is $I = \frac{12}{6 + 4} = 1.2 A$
Potential across $R_{1}$ is $V_{1} = I R_{1} = 1.2 \times 6 = 7.2 V$ and
potential across $R_{2}$ is $V_{2} = I R_{2} = 1.2 \times 4 = 4.8 V$

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