A lamp is connected to a battery containing two cells as shown - Scottish Highers Physics - Question 12 - 2017

To calculate the lost volts, we use the formula:

$V_{lost} = I imes r$

Where:

• $I = 64 imes 10^{-3}$ A (current from ammeter)
• $r = 2 imes 2$ Ω (total internal resistance of two cells)

Calculating: $V_{lost} = 64 imes 10^{-3} imes (2 imes 2) = 0.35 ext{ V}$

The voltmeter reading can be found using the formula:

$V = E - V_{lost}$

Where:

• $E = 1.5 imes 2$ V (total e.m.f. from two cells)

So:

• $E = 3.0$ V
• Thus, $V = 3.0 ext{ V} - 0.35 ext{ V} = 2.65 ext{ V}$

The reading on the voltmeter is 2.65 V.

To find the power dissipated, we use:

$P = I^2 imes R$

Where:

• $I = 64 imes 10^{-3}$ A
• $R = 2 ext{ Ω}$ (the resistance of the lamp)

Calculating: $P = (64 imes 10^{-3})^2 imes 2 = 0.008192 imes 2 = 0.016384 ext{ W}$

Converting to a more significant figure: The power dissipated by the lamp is approximately 0.017 W.

Using the formula that relates e.m.f., lost volts, and the current through the circuit,

$V = I(R + r)$

Where:

• $V = 3.6$ V (potential difference across LED)
• $I = 26 imes 10^{-3}$ A
• $r = 2 imes 4$ Ω (total internal resistance of four cells)

Solving for R:

1. Substitute the values: $3.6 = 26 imes 10^{-3} (R + 8)$
2. Isolate R: $R = \frac{3.6}{26 imes 10^{-3}} - 8$
3. Calculate the values: $R = 138.46 - 8 = 130.46 ext{ Ω}$

Therefore, the resistance of R is approximately 82.0 Ω.