Q160 / 180 · COMEDK 2025 (10 May Shift 2) · Physics
The ratio of the number of turns of the primary coil to the secondary coil of an ideal transformer is $5: 1$. The primary is connected to 220 V supply. What is the output power of the transformer if the output resistance is $11 \Omega$ ?
- A1100 W
- B100 W
- Correct answerC176 W
- D1760 W
Solution
For an ideal transformer, the relation between the primary voltage $V_p$, secondary voltage $V_s$, and the number of turns $N_p$ and $N_s$ is given by $\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$.
Given $\dfrac{N_p}{N_s} = \dfrac{5}{1}$, we have $\dfrac{N_s}{N_p} = \dfrac{1}{5}$.
Given $V_p = 220$ V, the secondary voltage is $V_s = V_p \times \dfrac{N_s}{N_p} = 220 \times \dfrac{1}{5} = 44$ V.
The output power $P_s$ is given by $P_s = \dfrac{V_s^2}{R_s}$, where $R_s = 11 \Omega$ is the output resistance.
Substituting the values, $P_s = \dfrac{44^2}{11} = \dfrac{1936}{11} = 176$ W.
Answer: 176 W
Given $\dfrac{N_p}{N_s} = \dfrac{5}{1}$, we have $\dfrac{N_s}{N_p} = \dfrac{1}{5}$.
Given $V_p = 220$ V, the secondary voltage is $V_s = V_p \times \dfrac{N_s}{N_p} = 220 \times \dfrac{1}{5} = 44$ V.
The output power $P_s$ is given by $P_s = \dfrac{V_s^2}{R_s}$, where $R_s = 11 \Omega$ is the output resistance.
Substituting the values, $P_s = \dfrac{44^2}{11} = \dfrac{1936}{11} = 176$ W.
Answer: 176 W
Hence, the correct answer is option: 'C'.