In the circuit shown below

If each thyristor voltage drop is $1.5 V$ while conducting, then the required peak inverse voltage of each thyristor is

238.5 V

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118.5 V

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170 V

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168.2 V

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Solution

The correct option is D $168.2 V$
When $T_{1}$ is open and $T_{2}$ is conducting,
At time instant when $V_{s} = - V_{m}$ , let us apply KVL in the loop shown below,

PIV across $T_{1} = V_{m} - 1.5$
$= 120 \sqrt{2} - 1.5 = 168.2 V$

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In the circuit shown below If each thyristor voltage drop is 1.5V while conducting, then the required peak inverse voltage of each thyristor is

The correct option is D 168.2VWhen T1 is open and T2 is conducting, At time instant when Vs=−Vm, let us apply KVL in the loop shown below, PIV across T1=Vm−1.5 =120√2−1.5=168.2V

In the circuit shown below

If each thyristor voltage drop is $1.5 V$ while conducting, then the required peak inverse voltage of each thyristor is

The correct option is D $168.2 V$
When $T_{1}$ is open and $T_{2}$ is conducting,
At time instant when $V_{s} = - V_{m}$ , let us apply KVL in the loop shown below,

PIV across $T_{1} = V_{m} - 1.5$
$= 120 \sqrt{2} - 1.5 = 168.2 V$