In the circuit shown below- if the resistance 5 - develops a heat 42J per second- then heat developed in 2 - must be about-A- 30 J - sB- 20 J - sC- 25 J - sD- 35 J - s

The correct option is A 30 J/s From diagram I_1 (6+9)=I_2 (5) Now, By KCL I=I_1+I_2=4I_2 /3 P_1 =I_2^2(5)=42 ...(i) P_2 =I^2(2)=16/9I_2^2 (2) =16/9×42/5× 2≈ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Kirchhoff's Voltage Law

In the circui...

Question

In the circuit shown below, if the resistance $5 Ω$ develops a heat 42J per second, then heat developed in $2 Ω$ must be about:

20 J/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25 J/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30 J/s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

35 J/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 30 J/s

From diagram
$I_{1} (6 + 9) = I_{2} (5)$
Now, By KCL
$I = I_{1} + I_{2} = \frac{4 I_{2}}{3}$
$P_{1} = I_{2}^{2} (5) = 42$ ...(i)
$P_{2} = I^{2} (2) = \frac{16}{9} I_{2}^{2}$ (2)
$= \frac{16}{9} \times \frac{42}{5} \times 2 \approx J / s$

Suggest Corrections

Similar questions

Q. In the circuit shown in figure, the 5

Ω

resistance develops 20.00 cal/s due to the current flowing through it. The heat developed in 2

Ω

resistance (in cal/s) is

Q. Find total heat developed per second in the circuit.

Q. How many joules of electrical energy are transferred per second by a 6 V; 0.5 A lamp?

(a) 30 J/s
(b) 12 J/s
(c) 0.83 J/s
(d) 3 J/s

Q. In the circuit,

5 Ω

resistor develops 45J/s due to current flowing through it. The power developed per second across

12 Ω

resistor is:

Q. A force

F = 10 N

is acting on pulley of mass

M = 2 kg

and radius

R = 1 m

as shown in the figure. If pulley is rotating with constant angular velocity

ω = 4 rad/s

. Find the power delivered by the force

F

to the pulley.

Join BYJU'S Learning Program

In the circuit shown below, if the resistance 5Ω develops a heat 42J per second, then heat developed in 2Ω must be about:

The correct option is C 30 J/s From diagram I1(6+9)=I2(5) Now, By KCL I=I1+I2=4I23 P1=I22(5)=42 ...(i) P2=I2(2)=169I22 (2) =169×425×2≈J/s

In the circuit shown below, if the resistance $5 Ω$ develops a heat 42J per second, then heat developed in $2 Ω$ must be about:

The correct option is C 30 J/s

From diagram
$I_{1} (6 + 9) = I_{2} (5)$
Now, By KCL
$I = I_{1} + I_{2} = \frac{4 I_{2}}{3}$
$P_{1} = I_{2}^{2} (5) = 42$ ...(i)
$P_{2} = I^{2} (2) = \frac{16}{9} I_{2}^{2}$ (2)
$= \frac{16}{9} \times \frac{42}{5} \times 2 \approx J / s$