KCET · Physics · Current Electricity
Three resistors \(1 \Omega, 2 \Omega\) and \(3 \Omega\) are connected to form a triangle. Across \(3 \Omega\) resistor a \(3 \mathrm{~V}\) battery is connected. The current through \(3 \Omega\) resistor is
- A \(0.75 \mathrm{~A}\)
- B \(1 \mathrm{~A}\)
- C \(2 \mathrm{~A}\)
- D \(1.5 \mathrm{~A}\)
Answer & Solution
Correct Answer
(B) \(1 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
Required arrangement is shown in figure.
The equivalent circuit will look like (since the two resistances of \(1 \Omega\) and \(2 \Omega\) are in series, which form \(3 \Omega\) which is in parallel with \(3 \Omega\) resistance).
Therefore, the effective resistance is
\(\frac{(1+2) \times 3}{(1+2)+3}=\frac{3}{2} \Omega\)
\(\therefore\) Current in the circuit,
\(I=\frac{3}{(3 / 2)}=2 \mathrm{~A}\)
\(\therefore\) Current in \(3 \Omega\) resistor \(=\frac{\mathrm{I}}{2}=1 \mathrm{~A}\)
The equivalent circuit will look like (since the two resistances of \(1 \Omega\) and \(2 \Omega\) are in series, which form \(3 \Omega\) which is in parallel with \(3 \Omega\) resistance).
Therefore, the effective resistance is
\(\frac{(1+2) \times 3}{(1+2)+3}=\frac{3}{2} \Omega\)
\(\therefore\) Current in the circuit,
\(I=\frac{3}{(3 / 2)}=2 \mathrm{~A}\)
\(\therefore\) Current in \(3 \Omega\) resistor \(=\frac{\mathrm{I}}{2}=1 \mathrm{~A}\)
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