Question

refer to the circuit shown. what will be the total power dissipation in the circuit if P is the power dissipated in R1 ? It is given that R2 = 4 R1 and R3 = 12 R1

Answer 1

Given:

The resistance given R2 =2R1

The resistance R3= 12 R1

To find:

The resistance of first is R1

Solution:

The total equivalent resistance be

=R2.R3/R2+R3  + R1

Putting the above values we get

3R1 + R1 =4R1

The power of dissipated be

P=R1=I²R1

Putting the above value

=4I²R1

=4P

The power of dissipation be 4P

Answer 2

Given:

• A circuit is given.

To Find:

• Power dissipation in the circuit

Solution:

Resistance $R_2 \ and \ R_3$  are in parallel so equivalent resistance is given by

                     $R_P = \frac{R_2 \times R_3 }{R_2 + R_3}$

      ∴  $R_2 = 4 R_1 \ and \ R_3 = 12 R_1$

                    $R_{p} = \frac{4 R_1 \times 12 R_1 }{4 R_1 + 12 R_1}$

                   $R_{p} = \frac{48 R_1 ^2 }{ 16 R_1}$

                 $R_{p} = 3 R_1$

Total equivalent resistance of the circuit

                   $R_{eq} = R_P + R_1$

                   $R_{eq} = 3 R_1 + R_1 = 4R_1$

Power dissipation at $R_1$

                  $P = I^2 R_1$

Total Power dissipation in circuit

            $P_{total} = I^2 R_{eq}$

          $P_{total} = I^2 \times 4R_1$

          $P_{total} = 4 \times I^2R_1$

           $P_{total} = 4P$

Thus, Total Power dissipation in circuit is $4P$