Consider the circuit of three identical batteries and three identical bulbs conn

Question

Consider the circuit of three identical batteries and

three identical bulbs connected exactly as shown

here. In this problem, assume that each battery is a

perfect physics battery supplying voltage V and the

bulbs are perfect ohmic devices with resistance R.

A friend of yours says, That middle battery is

hooked up backwards! The bulbs wont light if you

hook the circuit up like that.

(a) Respectfully and thoroughly explain to your

friend that the bulbs will indeed light.

In your explanation, use both math and

conceptual understanding. Can you use the

ski-area analogy to help? (This involves using a ski resort as an analogy to your potential energy as you go around a circuit. If you go up a lift, you have more PE, when you ski down, you lose your PE, etc.)

(b) Find the power dissipated by each bulb and the total power dissipated by the circuit.

Consider the circuit of three identical batteries and three identical bulbs connected exactly as shown here. In this problem, assume that each battery is a perfect physics battery supplying voltage V and the bulbs are perfect ohmic devices with resistance R. A friend of yours says, That middle battery is hooked up backwards! The bulbs wont light if you hook the circuit up like that. (a) Respectfully and thoroughly explain to your friend that the bulbs will indeed light. In your explanation, use both math and conceptual understanding. Can you use the ski-area analogy to help? (This involves using a ski resort as an analogy to your potential energy as you go around a circuit. If you go up a lift, you have more PE, when you ski down, you lose your PE, etc.) (b) Find the power dissipated by each bulb and the total power dissipated by the circuit.

Explanation / Answer

a)

here , as if we start skiing at the bottom towrds the up in the left side

then ,

here it will go down two batteries and go up only one battery ,

it will stil have kinetic energy and will keep on moving .

hence , the current will flow in the current and light bulbs will glow

Now , Using KVL , let the current in the circuit is

V - V + V -iR -iR -iR =0

i = V/3R

as the current is non-zero , the light buls will glow.

b)

Power dissipated by each bulb , Pb = i^2*R

Pb = (V/3R)^2 * R

Pb = V^2/9R

the power dissipated by 1 bulb is V^2/9R

Total Power = 2*Pb

Total Power = V^2/9R * 3

Total Power = V^2/3R

the total power is V^2/3R

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