Question

in the circuit shown the thermal power dissipated in R1 is p the thermal power dissipated in R2 is​

Answer 1

Answer:

The thermal power dissipated in R2 = 4/9 × P or 4P/9.

Explanation:

Given;

R1 = R

R2 = R

R3 = 2R

Power dissipated in R1 = P

Formula to be used : Power = Potential difference(V) × Current(I)

Ohm's law (V = IR)

Let R2 and R3 be Rp.

thus Rp equals;

1/Rp= 1/R2 + 1/R3

1/Rp = 1/R + 1/2R

1/Rp = 3/2R

Rp = 2/3 R;

Now let the current in circuit be i.

Thus, current in Rp and R1 will be same since they are in series.

Potential difference in Rp = i × (2/3 R)

= 2/3 × iR

Here, potential difference will be same in both R2 and R3.

Thus, p.d. in R2 = 2/3 × iR

Now, current in R2 = P.d or potential difference / R2

hence, I or current in R2 = (2/3×iR) / R

= 2/3×i or 2i/3

Thus power dissipated by R2 = p.d. in R2 × I in R2

= 2/3 × iR × 2/3 × i

= 4/9 i^2R. (i)

Now, it is given that power dissipated in R1 = P.

which means, P = iV

= i^2R1 (V = iR1)

= i^2R. (R1 = R) (ii)

From equation (i) and (ii), it can be deduced that;

Power dissipated in R2 = 4/9 ×P. (i^2/R = P)

Thus, option 2 is the right one.

Thats all.