Fig. 3.1 shows a circuit consisting of a battery of electromotive force 16.0V and negligible internal resistance, two resistors and a thermistor - OCR - A-Level Physics A - Question 3 - 2013 - Paper 1

Internal resistance refers to the resistance within the battery itself that opposes the flow of current. It results in a voltage drop when current flows, thereby reducing the effective voltage available to the circuit.

To find the total resistance R, we sum the resistance of the thermistor and the 120Ω resistor:

$R = R_{thermistor} + R_{120} = 360Ω + 120Ω = 480Ω$

Using Ohm's Law, the potential difference across the thermistor can be calculated using the total voltage provided by the battery:

The current I in the circuit can be calculated as: $I = \frac{E}{R} = \frac{16.0V}{480Ω} = 0.0333A$

Then, using Ohm's Law again, the potential difference V across the thermistor is: $V = I \times R_{thermistor} = 0.0333A \times 360Ω = 12V$

As the temperature increases, the resistance of the thermistor (which is a temperature-dependent component) typically decreases. This leads to an increase in current through the circuit due to Ohm's Law. Since voltage is constant, the potential difference across the thermistor will also decrease as its resistance decreases and the current increases, resulting in a greater proportion of the total voltage being dropped across the fixed resistor instead.

The charge Q stored in the battery can be calculated using the formula: $Q = I \times t$ Where:

• I is the current (1.2A)
• t is the time in seconds (8 hours = 8 \times 3600 = 28800 seconds)

Thus, $Q = 1.2A \times 28800s = 34560C$

Using the power loss rate and the total energy stored in the battery:

Power loss = 1.4 J/s Time taken to discharge can be calculated as follows: First calculate total energy: $Energy = Q \times E = 34560C \times 16V = 552960J$ Then, $Time = \frac{Energy}{Power} = \frac{552960J}{1.4J/s} = 394257.14 seconds\approx 109.5 hours$