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Chapter 4: Q12DQ (page 873)

Consider the circuit shown in Fig. Q26.12. What happens to the brightness of the bulbs when the switch S is closed if the battery

(a) has no internal resistance and

(b) has nonnegligible internal resistance? Explain why.

Short Answer

Expert verified
  1. The brightness will remain the same.

b. The brightness will decrease.

Step by step solution

01

Parallel Connection

not given in the drive

02

When there is no internal resistance

  1. When S is closed, the bulbs are in parallel and have the same voltage drop because they have no internal resistance (r),

ε=V+Ir

r = 0 then,

ε=V

The voltage will remain constant as the equivalent resistance R and current I change, and in this case, the voltage drop is the same for all three branches because they are connected in parallel.

Hence, the brightness will remain the same.

03

When there is an internal resistance that is non-negligible

a. As the resistance (r) has a value, therefore, the voltage drop in the circuit is:

ε=V+lr

Because the equivalent resistance reduces, the current increases, lowering the voltage drop across the battery due to internal resistance Ir.

Therefore, the brightness decreases as the voltage drop decreases.

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