Find maximum voltage across AB in the circuit shown in figure. Assume that the diode is ideal.

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 `15k Omega ` and the effective resistance across AB is
`R_(AB ) = ( 10 xx 15 )/( 10 +15 )=( 10 xx 15 )/( 25 ) = 6 k Omega `
` 6 k Omega `is in series with `5k Omega `
` therefore ` total resistance = `R_T = 6k Omega +5k Omega =11k Omega `
V=30V. 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 6 k Omega = 16.32 V `