Question 23
A battery with e.m.f. E and
internal resistance r is connected in series with a variable external
resistor.
The value of the
external resistance R is slowly increased from zero.
Which statement is
correct? (Ignore any temperature effects.)
A The
potential difference across the external resistance decreases.
B The
potential difference across the internal resistance increases.
C The
power dissipated in r increases and then decreases.
D The power dissipated in R increases
and then decreases.
Reference: Past Exam Paper – June 2015 Paper 11 Q36
Solution:
Answer: D.
Ohm’s law: V = IR
In general, as the resistance of a component increases
the p.d. across it increases.
So, as the value of external resistance R is
increased, the p.d. across it increases while the p.d. across the internal
resistance decreases (since the sum of p.d. in a circuit is equal to the e.m.f.
in the circuit). [A and B are incorrect]
The external resistance R and the internal resistance
r forms a potential divider.
Power dissipated in external resistance R = I2R
Initially, the value of R is zero, so the power
dissipated across it is also zero.
Similarly, if the value of R is infinite, the power
dissipated is also zero (as the current I will then be zero).
So, the power dissipated must increase and then
decrease as R increases from zero. As the value of R increases, the current I
decreases, the p.d. across the external resistance R increases. The p.d. across
the internal resistance r decreases and the power dissipated in r (given by I2r)
decreases.
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