Find the maximum voltage across `AB` in the circuit shown in Fig. Assume that diode is ideal.
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Explanation

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`.