CUET Preparation Today
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$\text{From the figure the P.D across }2k\Omega \text{ resistor is 10V>}$ $I_L = \frac{V_z}{R_L} = \frac{10V}{2k\Omega} = 5mA$ $\text{P.D. across }4 K\Omega \text{ is 50V}$ $\Rightarrow I = \frac{50V}{4000\Omega} = 12.5mA$ $\Rightarrow I_Z = I - I_L = 7.5mA$ |
