Question

The network shown below is activated by a source of emf E and of internal resistance 2$\Omega$. If the current in 4 $\Omega$ resistor is 0.5 A, then,

• A. the emf of the source is 8 V.
• B. the current through 2 $\Omega$ is 2.75 A.
• C. the potential difference between A and B is 8 V.
• D. the current through 6 $\Omega$ resistor is 1.33 A.

Answer 1

Answer: (B), (C), (D)

The correct options are
B the current through 2 $\Omega$ is 2.75 A.
C the current through 6 $\Omega$ resistor is 1.33 A.
D the potential difference between A and B is 8 V.

P.D. across 4 $\Omega =0.5\times 4=2V$
$\Rightarrow$ current through 8 $\Omega$parallel $=\frac{2}{8}=0.25A$
Hence$i_1=0.5+0.25=0.75A$
P.D. between A and B$=2+8\times \frac{3}{4}=8V$
Now $i_2=\frac{8\times 18}{6\times 12}=2A$
Hence current through $2\Omega =i_1+i_2=2.75A$
Further, $V_A-E+2\times 2.75=V_B$
$E=(V_A-V_B)+5.50=8+5.5=13.5V$
the current through 6 $\Omega$ is $i_2\times \frac{12}{(12+6)}=\frac{2\times 12}{18}=\frac{4}{3}=1.33A$.
So, B, C, D are correct.