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Find maximum voltage across AB in the circuit shown in Fig. Assume that diode is ideal. |
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Answer» Solution :As the DIODE is treated ideal, its forward RESISTANCE `R_(f)=` zero. It acts as short circuit. So, `10 kOmega` is in parallel with `15kOmega` and the effective resistance across AB is `R_(AB)=(10xx15)/(10+15)=(10xx15)/(25)=6kOmega` `6kOmega` in series with `5kOmega`. `therefore` total resistance `=R_(T)=6kOmega+5kOmega=11kOmega` `V=30V` Current drawn from the battery is `I=(V)/(R_(T))=(30V)/(11kOmega)=2.72mA` `V_(AB)=IR_(AB)=2.72mA xx6kOmega=16.32V` |
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