Current Electricity Ques 39
39. The current (I) in the given circuit is
(a) $1.6 A$
(b) $2.0 A$
(c) $0.32 A$
(d) $3.2 A$
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Answer:
Correct Answer: 39.(b)
Solution:
- (b) In circuit, $R_B$ and $R_C$ are in series, so, $R_s=$ $6+6=12 \Omega$. This $12 \Omega$ resistance is in parallel with $R_A=3 \Omega$,
So, equivalent resistance of circuit
$R=\frac{3 \times 12}{3+12}=\frac{36}{15}=\frac{12}{5} \Omega$
$\therefore$ Current in circuit, $I=\frac{V}{R}$
$=\frac{4.8}{(\frac{12}{5})}=\frac{4.8 \times 5}{12}=2 A$
BETA
