Q147 / 160 · AP EAMCET 2023 (16 May Shift 1) · Physics
In the circuit given below, if the bulb is to glow with maximum intensity, the value of ' $R$ ' is
(neglect internal resistance of the cell)
- Correct answerA$1.25 \Omega$
- B$4.5 \Omega$
- C$6 \Omega$
- D$8.5 \Omega$
Solution
Resistance of bulb, $R=\frac{V^2}{P}$
$=\frac{(1.5)^2}{0.45}=5 \Omega$
$P=i_1^2 R$ $\begin{aligned} & 0.45=\mathrm{i}_1{ }^2 5 \\ & \mathrm{i}_1=0.3 \mathrm{~A}\end{aligned}$ Apply KVL, We have $\begin{aligned} & -6+3 \mathrm{i} 1.5=0 \\ & \mathrm{i}=\frac{6-1.5}{3}=1.5 \mathrm{~A} \\ & 1.5=\left(\mathrm{i}-\mathrm{i}_1\right) \mathrm{R} \\ & 1.5=(1.5-0.3) \mathrm{R} \\ & \mathrm{R}=\frac{1.5}{1.2}=1.25 \Omega\end{aligned}$
$P=i_1^2 R$ $\begin{aligned} & 0.45=\mathrm{i}_1{ }^2 5 \\ & \mathrm{i}_1=0.3 \mathrm{~A}\end{aligned}$ Apply KVL, We have $\begin{aligned} & -6+3 \mathrm{i} 1.5=0 \\ & \mathrm{i}=\frac{6-1.5}{3}=1.5 \mathrm{~A} \\ & 1.5=\left(\mathrm{i}-\mathrm{i}_1\right) \mathrm{R} \\ & 1.5=(1.5-0.3) \mathrm{R} \\ & \mathrm{R}=\frac{1.5}{1.2}=1.25 \Omega\end{aligned}$
Hence, the correct answer is option: 'A'.