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GUJCET · Physics · Current Electricity
The total current supplied to the circuit by the battery is _________.

- A 6 A
- B 4 A
- C 2 A
- D 1.5 A
Answer & Solution
Correct Answer
(D) 1.5 A
Step-by-step Solution
Detailed explanation
D
Here \(6 \Omega\) and \(2 \Omega\) resistors are in parallel.
so equivalent resistance
\(\begin{array}{l}R_1=\frac{6 \times 2}{6+2} \\ R_1=\frac{12}{8}\end{array}\)

\(R_{I}=1.5 \Omega\)
\(R _1=1.5,3 \Omega\) and \(1.5 \Omega\) are in series connection. So equivalent resistance
\(\begin{array}{l} R _2= R _1+1.5+3 \\ R _2=1.5+1.5+3 \\ R _2=6 \Omega\end{array}\)
current \(I=\frac{V}{R}\)
\(\begin{array}{l}I=\frac{9}{6} \\ I=1.5 A\end{array}\)
Here \(6 \Omega\) and \(2 \Omega\) resistors are in parallel.
so equivalent resistance
\(\begin{array}{l}R_1=\frac{6 \times 2}{6+2} \\ R_1=\frac{12}{8}\end{array}\)

\(R_{I}=1.5 \Omega\)
\(R _1=1.5,3 \Omega\) and \(1.5 \Omega\) are in series connection. So equivalent resistance
\(\begin{array}{l} R _2= R _1+1.5+3 \\ R _2=1.5+1.5+3 \\ R _2=6 \Omega\end{array}\)
current \(I=\frac{V}{R}\)
\(\begin{array}{l}I=\frac{9}{6} \\ I=1.5 A\end{array}\)
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