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The current drawn from the primary of a transformer which steps down 220 V to 22 V at 0.1 amp. shall be |
1A 0.1A 0.2A 0.01A |
0.01A |
Here np = 220 V, ns = 22 V, Is = 0.1 amp. and Ip = ? $\frac{E_s}{E_p} = \frac{I_p}{I_s} = \frac{n_s}{n_p}$ $\Rightarrow I_p = \frac{E_s}{E_p} \times I_s = \frac{22}{220} \times 0.1A = 0.01A$ |
