A potential divider circuit consists of a light-dependent resistor (LDR) connected in series with a fixed resistor of resistance 1.0 k. This series combination is connected across a battery of e.m.f. 9.0 V and negligible internal resistance. The output potential difference Vout…

A Level Physics 9702: Potential Divider with an LDR

A potential divider circuit consists of a light-dependent resistor (LDR) connected in series with a fixed resistor of resistance $1.0\text{ k}\Omega$ . This series combination is connected across a battery of e.m.f. $9.0\text{ V}$ and negligible internal resistance. The output potential difference $V_{\text{out}}$ is taken across the fixed resistor.

In darkness, the resistance of the LDR is $8.0\text{ k}\Omega$ .

(a) State how the resistance of an LDR changes as the light intensity falling on it increases. [1]

(b) Calculate $V_{\text{out}}$ in darkness. [2]

(c) The LDR is illuminated with bright light, causing its resistance to fall to $1.0\text{ k}\Omega$ . Calculate the new value of $V_{\text{out}}$ . [2]

(d) State and explain how $V_{\text{out}}$ changes as the light intensity increases, and suggest one practical device in which this potential divider circuit could be used. [3]

Part (a): How LDR resistance changes with light intensity

The resistance of a light-dependent resistor (LDR) decreases as the light intensity falling on it increases, because more photons free more charge carriers in the semiconductor material, allowing charge to flow more easily.

Part (b): Output p.d. in darkness

In a potential divider, the p.d. across a component is in proportion to its share of the total resistance. Here $V_{\text{out}}$ is taken across the fixed resistor: $V_{\text{out}} = V_{\text{supply}}\times\frac{R_{\text{fixed}}}{R_{\text{fixed}}+R_{\text{LDR}}}$

Substituting $V_{\text{supply}}=9.0\text{ V}$ , $R_{\text{fixed}}=1.0\text{ k}\Omega$ , $R_{\text{LDR}}=8.0\text{ k}\Omega$ : $V_{\text{out}} = 9.0\times\frac{1.0}{1.0+8.0} = 9.0\times\frac{1.0}{9.0} = 1.0\text{ V}$

Check using the other component instead: the p.d. across the LDR is $9.0\times\frac{8.0}{9.0}=8.0\text{ V}$ , and $1.0+8.0=9.0\text{ V}$ , matching the supply e.m.f. Confirmed.

Part (c): Output p.d. in bright light

The LDR’s resistance has fallen to $1.0\text{ k}\Omega$ , so the total resistance is now $1.0+1.0=2.0\text{ k}\Omega$ : $V_{\text{out}} = 9.0\times\frac{1.0}{1.0+1.0} = 9.0\times\frac{1.0}{2.0} = 9.0\times0.50 = 4.5\text{ V}$

Check using decimals for the fraction: $\frac{1.0}{2.0}=0.50$ exactly, so $V_{\text{out}}=9.0\times0.50=4.5\text{ V}$ . Same result both ways.

Part (d): Trend and application

As the light intensity increases, the LDR’s resistance falls (part (a)), so it takes a smaller share of the fixed $9.0\text{ V}$ supply, leaving a larger share for the fixed resistor. Since $V_{\text{out}}$ is taken across the fixed resistor, $V_{\text{out}}$ increases as light intensity increases, consistent with the rise from $1.0\text{ V}$ to $4.5\text{ V}$ found in parts (b) and (c).

This kind of circuit, where the output p.d. rises with light intensity, is used in devices such as a light-operated switch or automatic street lamp control: because the sensor should switch a lamp on when it goes dark (i.e. when $V_{\text{out}}$ is low), the low-light output of this arrangement can be compared against a fixed reference voltage to trigger the lamp at dusk.

Final answers

(a) LDR resistance $\boxed{\text{decreases}}$ as light intensity increases
(b) $V_{\text{out}} = \boxed{1.0}\text{ V}$ in darkness
(c) $V_{\text{out}} = \boxed{4.5}\text{ V}$ in bright light
(d) $V_{\text{out}}$ increases as light intensity increases; usable in a light-operated switch or automatic street lamp control