
The diagram shows a circuit containing a cell, an ammeter, a lamp and a variable resistor. The resistance of the variable resistor is increased. What happens to the ammeter reading and what happens to the brightness of the lamp.
Ammeter reading Lamp brightness A. Decreases Decreases B Decreases Increases C Increases Decreases D. Increases Increases

Ammeter reading | Lamp brightness | |
A. | Decreases | Decreases |
B | Decreases | Increases |
C | Increases | Decreases |
D. | Increases | Increases |
Answer
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Hint: We need to investigate the current in the circuit using Kirchhoff’s voltage rule. The reduction of power supplied to the lamp corresponds to decrease in brightness, and increase in power causes increase in brightness. Using this we can find the correct answer.
Formula used: In this solution we will be using the following formulae;
$ P = IV $ where $ P $ is the power supplied to a load such as a lamp, $ I $ is the current flowing through the load, and $ V $ is the potential difference across it.
$ \sum V = 0 $ where $ V $ is the potential difference across each element of the loop.
Complete step by step solution:
A variable resistor is simply a resistor whose resistance value can easily be changed. The voltage drop across it is as in a normal resistor. Hence, on applying Kirchhoff’s voltage law across the loop, we have
$ E - Ir - IR = 0 $ where $ E $ is the voltage value of the cell, $ r $ is the resistance due to the lamp, and $ R $ is the resistance value of the variable resistor.
The equation by factoring $ I $ can become
$ E - I(r + R) = 0 $
To investigate the effect of $ R $ on $ I $ we must make $ I $ subject to the formula. Hence
$ E = I(r + R) $
$ \Rightarrow I = \dfrac{E}{{r + R}} $
This informs us that an increase in the value of resistance $ R $ reduces the current $ I $ .
Also, from $ P = IV $ we see that if the current reduces, the power consumed by the lamp reduces, thus, the brightness also decreases.
Therefore, when the resistance of the variable resistor $ R $ increases, the current reduces and thus, the brightness reduces
Hence, the correct option is A.
Note:
In application, variable resistors are used in situations where resistance in a circuit needs to be easily reduced. A few appliances that use variable resistors include volume dials of radios and standing fan controllers.
Formula used: In this solution we will be using the following formulae;
$ P = IV $ where $ P $ is the power supplied to a load such as a lamp, $ I $ is the current flowing through the load, and $ V $ is the potential difference across it.
$ \sum V = 0 $ where $ V $ is the potential difference across each element of the loop.
Complete step by step solution:
A variable resistor is simply a resistor whose resistance value can easily be changed. The voltage drop across it is as in a normal resistor. Hence, on applying Kirchhoff’s voltage law across the loop, we have
$ E - Ir - IR = 0 $ where $ E $ is the voltage value of the cell, $ r $ is the resistance due to the lamp, and $ R $ is the resistance value of the variable resistor.
The equation by factoring $ I $ can become
$ E - I(r + R) = 0 $
To investigate the effect of $ R $ on $ I $ we must make $ I $ subject to the formula. Hence
$ E = I(r + R) $
$ \Rightarrow I = \dfrac{E}{{r + R}} $
This informs us that an increase in the value of resistance $ R $ reduces the current $ I $ .
Also, from $ P = IV $ we see that if the current reduces, the power consumed by the lamp reduces, thus, the brightness also decreases.
Therefore, when the resistance of the variable resistor $ R $ increases, the current reduces and thus, the brightness reduces
Hence, the correct option is A.
Note:
In application, variable resistors are used in situations where resistance in a circuit needs to be easily reduced. A few appliances that use variable resistors include volume dials of radios and standing fan controllers.
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