A variable resistor is used to control the current in a circuit, as shown in Fig. 5.1.

Question 35

(a) A variable resistor is used to control the current in a circuit, as shown in Fig. 5.1.

 Fig. 5.1

The variable resistor is connected in series with a 12 V power supply of negligible internal resistance, an ammeter and a 6.0 Ω resistor. The resistance R of the variable resistor can be varied between 0 and 12 Ω.

(i) The maximum possible current in the circuit is 2.0 A. Calculate the minimum possible current. [2]

(ii) On Fig. 5.2, sketch the variation with R of current I1 in the circuit.

Fig. 5.2

[2]

(b) The variable resistor in (a) is now connected as a potential divider, as shown in Fig. 5.3.

Fig. 5.3

Calculate the maximum possible and minimum possible current I2 in the ammeter. [2]

(c) (i) Sketch on Fig. 5.4 the I – V characteristic of a filament lamp.

Fig. 5.4

[2]

(ii) The resistor of resistance 6.0 Ù is replaced with a filament lamp in the circuits of Fig. 5.1 and Fig. 5.3. State an advantage of using the circuit of Fig. 5.3, compared to the circuit of Fig 5.1, when using the circuits to vary the brightness of the filament lamp. [1]

Reference: Past Exam Paper – June 2011 Paper 21 Q5

Solution:

(a)

(i)

{The variable resistor and the fixed resistor are connected in series. So, the same current flows through them.

Current is maximum when the resistance of the variable resistor is minimum and current is minimum when the resistance of the variable resistor is maximum.

V = IR

I = V / R          where R is the total resistance}

Minimum current, I = 12 / (6 + 12) = 0.67 A

(ii)

correct start and finish points

correct shape for curve with decreasing gradient

{From part (i),

When R = 0, current is maximum and equal to 2.0 A. This is the point (0, 2.0).

When R is maximum, the current is 0.67 A. This is point (12, 0.67).

To obtain the shape of the graph (increasing or decreasing gradient between these 2 points), we need to calculate more point for the graph.

The value of the variable resistor R is changed and the new current is calculated.

I = 12 / (6+R)

When R = 6 Ω, I = 1.0 A

When R = 4 Ω, I = 1.2 A

When R = 8 Ω, I = 0.86 A

When R = 10 Ω, I = 0.75 A

When R = 2 Ω, I = 1.5 A

Note that the curve should not go beyond R = 12 Ω.}

(b)

maximum current = 2.0 A

minimum current = 0

{The variable resistor R is now connected in parallel to the fixed resistor. The total current from the battery splits at the junction connecting the fixed resistor and the variable resistor. I₂ is the current flowing through the 6.0 Ω resistor.

The greater the resistance of a component, the smaller the current flowing.

When R = 0 Ω (minimum), all current would flow through it as its resistance is negligible. The current through the fixed resistor would be zero. This is the minimum current. (The fixed resistor has been short-circuited.)

When R = 12 Ω (maximum), the current flowing through it would be minimum. This means that most of the current would flow through the fixed resistor.

Since this is a parallel circuit, the p.d. across the fixed resistor is the same as the e.m.f. of the battery.

Current through fixed resistor: I₂ = V / R = 12 / 6 = 2.0 A}

(c) (i) The graph is a smooth curve starting at (0, 0) with a decreasing gradient. The end section should not be horizontal

(ii) With the circuit of Fig 5.3 (parallel combination), the full range of current / p.d. is possible.

OR The currents / p.d. can go down to zero

{Compare the minimum and maximum values of current in each case.

When connected in series: min current = 0.67 A and max current = 2.0 A

When connected in parallel: min current = 0 A and max current = 2.0 A}