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Find the maximum voltage across `AB` in the circuit shown in Fig. Assume that diode is ideal. . |
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Answer» As the diode is treated ideal, its forward resistance `R_(f) = zero`. It acts as short circuit. So `10 k Omega` is in parallel with `15 k Omega` and the effective resistance across `AB` is `R_(AB) =(10 xx 15)/(10+15)=(10 xx15)/(25) =6k Omega` `6 k Omega` is in series with `5 kOmega` `:.` total resistance `= R_(T) = 6 k Omega + 5 k Omega = 11 k Omega`, `V=30 V`. Current drawn from the battery is `I=(V)/(R_(T))=(30V)/(11 k Omega) =2.72 mA` `V_(AB) =IR_(AB)=2.72 mA xx 6k Omega = 16.32 V`. |
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