Q122 / 180 · COMEDK 2015 · Physics
Several lamps of $50 \mathrm{~W}$ and $100 \mathrm{~V}$ rating are available. How many of them can be connected in parallel across a battery of a $120 \mathrm{~V}$ of internal resistance $10 \Omega$, so that all bulbs glow in full power?
- A2
- Correct answerB4
- C6
- D8
Solution
According to given condition, let $n$ number of bulbs are connected in parallel as shown below
$\begin{aligned} \text { Resistance of each bulb, } R &=\frac{V^{2}}{P}\left[\because P=\frac{V^{2}}{R}\right] \\ &=\frac{100^{2}}{50}=200 \Omega \end{aligned}$ Using KVL in the lowest loop, $$ \begin{array}{ll} & 120-10 I-\frac{I}{n} \times 200=0 \\ \Rightarrow \quad & 12-I=\frac{20 I}{n} \quad \text{...(i)} \end{array} $$ Power delivered by battery $=$ Sum of power used in bulbs $(120-10 I) I=50 n$ $\begin{array}{rlr}\Rightarrow & (12-I) I=5 n & \\ \Rightarrow & \frac{20 I}{n} \cdot I =5 n\end{array} \quad$ [from Eq. (i)] $\Rightarrow \quad 4 I^{2}=n^{2}$ $\Rightarrow \quad I=\frac{n}{2} \quad \text{...(ii)}$ From Eqs. (i) and (ii), we get $$ 12-\frac{n}{2}=\frac{20}{n} \times \frac{n}{2} \Rightarrow n=4 $$
$\begin{aligned} \text { Resistance of each bulb, } R &=\frac{V^{2}}{P}\left[\because P=\frac{V^{2}}{R}\right] \\ &=\frac{100^{2}}{50}=200 \Omega \end{aligned}$ Using KVL in the lowest loop, $$ \begin{array}{ll} & 120-10 I-\frac{I}{n} \times 200=0 \\ \Rightarrow \quad & 12-I=\frac{20 I}{n} \quad \text{...(i)} \end{array} $$ Power delivered by battery $=$ Sum of power used in bulbs $(120-10 I) I=50 n$ $\begin{array}{rlr}\Rightarrow & (12-I) I=5 n & \\ \Rightarrow & \frac{20 I}{n} \cdot I =5 n\end{array} \quad$ [from Eq. (i)] $\Rightarrow \quad 4 I^{2}=n^{2}$ $\Rightarrow \quad I=\frac{n}{2} \quad \text{...(ii)}$ From Eqs. (i) and (ii), we get $$ 12-\frac{n}{2}=\frac{20}{n} \times \frac{n}{2} \Rightarrow n=4 $$
Hence, the correct answer is option: 'B'.