Question

7. The electrical energy dissipated per second in a resistance of 422 is 100 J. The current flowing through the resistance will be (1) 25A (2) 15A (3) 10 A (4) 5 A ​

Answer 1

Answer:

2calorie/sec

Explanation:

Heat produced , H=I

2

Rt

here,

t

H

5

=I

5

2

R

5

=10⇒I

5

=

5

10

=

2

. This is the current through resistor 5Ω.

Voltage across resistor 5Ω is V

5

=I

5

R

5

=

2

×5=5

2

V

As resistors (4,6) and 5 are in parallel, so potential across (4,6) is equal to V

5

.

As resistor 4 and 6 are in series so current through each resistor I

4

=I

6

=

4+6

V

5

=

4+6

5

2

=

2

2

Heat produce in resistor 4 per sec =I

4

2

R

4

=(

2

2

)

2

×(4)=2cal/sec