The circuits shown below contain identical bulbs and non-identical ideal batteri

Question

The circuits shown below contain identical bulbs and non-identical ideal batteries. When a student sets up the circuits shown below, she observes that bulb A is brighter than bulb B. By this observation the student concludes that the voltage across bulb A is 1 greater than the voltage across bulb B. A second bulb is added to each circuit as shown below. The student notices that bulb A is now equal in brightness to bulb B. In terms of the voltage V2 across battery 2, the voltage V1 across battery 1 is 4V2 2V2 2V2 V2 V2/ 2 V2/2 V2/4 The batteries are now swapped, as below. The brightness of bulb A is now 3 less than the brightness of the bulb B.

Explanation / Answer

Power dissipated by the bulb = V^2 / R

where V and R are voltage across the bulb and resistance respectively.

Given, bulbs are identical. So, resistances of the bulbs are same.

In the first case, A is brighter than B. That means, power dissipated by A is more than B.

Since R is same, Voltage across A is greater than that of B

(b) If two resistors are in series, total voltage = sum of voltages across each resistor.

Since both the resistors are of same value, voltages acorss them are same in A.

So, voltage across each bulb in A = V1 /2

So, power dissipated by each bulb in A = (V1/2) 2 /R

= V12/4R

If resistors are in parallel, voltage across each resistor equals total voltage.

So, voltage across each bulb in B = V2

Power by each bulb in B = V22 /R

Given both powers are equal. So,

V12/4R = V22 /R

V1 = 2V2

(c) Voltage across bulb A = V2 /2

Power in A = V22/4R

Voltage across B = V1

Power in B = V12/R

Since V1 = 2V2 , power in B = 4V22/R, which is more than that in A.

So, brightness in A is less than that in B.

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