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In the circuit of figure. Treat the diod...

In the circuit of figure. Treat the diodes as ideal. Current in the `4` ohm resistor is

A

`2 A`

B

`3 A`

C

`12//7 A`

D

`30//13 A`

Text Solution

Verified by Experts

The correct Answer is:
A
With battery polarity, `D_(2)` is off. So, `I=12/(4+2)=2A`.
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Knowledge Check

  • A diode is a device that allows current to be carried in only one direction (the direction indicated by the arrowhead in its circuit symbol). Find (approximately) in terms of DeltaV and R the average power delivered to the diode circuit of the figure. Assume that the diodes are ideal and DeltaV is the rms voltage applied to the circuit.

    A
    `((DeltaV)^(2))/(R ) xx 0.6`
    B
    `((DeltaV)^(2))/(R ) xx 0.8`
    C
    `((DeltaV)^(2))/(R ) xx 1.1`
    D
    `((DeltaV)^(2))/(R ) xx 1.4`

  • In the circuit shown in fig the heat produced in the 5 ohm resistor due to the current flowing through it is 10 calories per second. The heat generated in the 4 ohms resistor is

    A
    `1cal//sec`
    B
    `2cal//sec`
    C
    `3cal//sec`
    D
    `4cal//sec`

  • In figure , assuming the diodes to be ideal ,

    A
    (a) ` D_1 ` is forward biased and ` D_2` is reverse biased and hence current flows from A to B.
    B
    ` D_2 ` is forward biased and `D_1` is reverse biased and hence no current flows from B to A and vice - versa.
    C
    `D_1` and `D_2` are both forward biased and hence current flows from A to B.
    D
    `D_1` and `D_2` are both reverse biased and hence no current flows from A to B and vice versa.
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