Calculate the voltmeter reading

[moenste]

Homework Statement

The light-dependent resistor (LDR) in the circuit below is found to have resistance 800 Ω in moonlight and resistance 160 Ω in daylight. Calculate the voltmeter reading, V_m, in moonlight with the switch S open.

If the reading of the voltmeter in daylight with the switch S closed is also equal to V_m what is the value of the resistance R?

Answers: 8.0 V, 50 Ω

2. The attempt at a solution
At first I would like to check whether I understand the scheme correctly. The switch S is a diode (open is a reverse-biased and closed is a forward-biased). The LDR is just a regular resistor, just need to take into account different values of resistance depending on the daytime. Should this be correct?

I this representation correct?

 

[gneill]

A switch is not a diode. A switch will pass current in either direction with equal facility and does not exhibit a potential drop when current passes through it (at least for an ideal switch). Diodes pass current in one direction only, and except for ideal diodes cause a potential drop on the order of a half volt when they do.

It's better to think of a switch as a zero resistance wire connection that can be inserted or removed manually.

Your redrawn circuit is fine.

 

[moenste]

A switch is not a diode. A switch will pass current in either direction with equal facility and does not exhibit a potential drop when current passes through it (at least for an ideal switch). Diodes pass current in one direction only, and except for ideal diodes cause a potential drop on the order of a half volt when they do.

It's better to think of a switch as a zero resistance wire connection that can be inserted or removed manually.

Your redrawn circuit is fine.

We need to find the voltmeter reading during moonlight when the resistance is 800 Ω and the switch is open. If the switch is open, the resistor R is shut off. Therefore we have only two resistors 200 + 800 = 1000 Ω. I = V / R = 10 / 1000 = 0.01 A. Now we find the 200 Ω has V₂₀₀ = I R₂₀₀ = 0.01 * 200 = 2 V, and the LDR resistor, as a result has 8 V. Since the voltmeter is connected to LDR, its reading is 8 V. Correct logic?

Update: for the second question we have: since the reading doesn't change, therefore V = I R and we can find I = V / R = 8 / 160 = 0.05 A of the whole circuit. Then, knowing V = 2 V and I = 0.05 A, we find the R_Parallel: R = V / I = 2 / 0.05 = 40 Ω. Then, by 1 / R = 1 / R + 1 / R → 1 / 40 = 1 / 200 + 1 / R → R = 50 Ω. Should be correct.

 

[gneill]

That all looks good. Well done.