Question

In the circuit shown in the figure, the current through

• A. the $3\Omega$ resistor is 0.50 A
• B. the $3\Omega$ resistor is 0.25 A
• C. the $4\Omega$ resistor is 0.50 A
• D. the $4\Omega$ resistor is 0.25 A

Answer 1

Answer: (D)

Net resistance of the circuit is $9\Omega.$
$\therefore $ Current drawn from the battery. = current through $3\Omega$ resistor
Potential difference between A and B is
$V_A-V_B=9-1(3+2)=4V=8i_1$
$\therefore i_1=0.5A$
$\therefore i_2=1-i_1=0.5A$
Similarly, potential difference between C and D
$V_C-V_D=(V_A-V_B)-i_2(2+2)$
$4-4i_2=4-4(0.5)=2V=8i_3$
$\therefore i_3=0.25A$
Therefore, $i_4=i_2-i_3=0.5-0.25$
$i_4=0.25A$