Fig. 3.9 shows an LDR connected in series with a 470 Ω resistor and a 6.0 V battery. The output to an alarm is taken across the 470 Ω resistor. In bright li...
Question 1 Report
Fig. 3.9 shows an LDR connected in series with a 470 Ω resistor and a 6.0 V battery. The output to an alarm is taken across the 470 Ω resistor. In bright light the LDR has a resistance of 200 Ω. In darkness its resistance rises to 15 kΩ.
Which statement about Vout is correct?
Answer Details
This circuit is a potential divider with the LDR on top and the fixed 470 \(\Omega\) resistor on the bottom. The output voltage \(V_{\text{out}}\) is taken across the 470 \(\Omega\) resistor:
\[ V_{\text{out}} = V_{\text{battery}} \times \frac{R_{\text{fixed}}}{R_{\text{LDR}} + R_{\text{fixed}}} \]
In bright light, the LDR resistance is low (200 \(\Omega\)), so the fixed resistor takes a large share of the total voltage:
\[ V_{\text{out}} = 6.0 \times \frac{470}{200 + 470} = 6.0 \times \frac{470}{670} \approx 4.2 \;\text{V} \]
In darkness, the LDR resistance is very high (15 000 \(\Omega\)), so almost all the voltage drops across the LDR and very little across the fixed resistor:
\[ V_{\text{out}} = 6.0 \times \frac{470}{15000 + 470} \approx 0.18 \;\text{V} \]
Therefore \(V_{\text{out}}\) is higher in bright light. The general principle is that in this arrangement, as the LDR resistance decreases (brighter light), a larger fraction of the supply voltage appears across the fixed resistor.
Download The App On Google Playstore
Everything you need to excel in your exams
