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What would this high resistance Voltmeter read and why? I have only just started A level physics so try to give answers relative to my level of understanding please!

The battery has a negligible internal resistance to be clear.

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Let's consider the circuit like a ideal generator of voltage $V_g$ connected to the two series of resistor. When you connect a generator to a series of resistors it behaves like a voltage divider and so the voltage across every resistor is proportional to its resistance. Hence, across the first resistor there will be a voltage of:

$$V_1 = V_g \frac{R_1}{R_1+R_2+R_3}$$

Similarly, across the fourth and fifth resistor there will be a voltage of: $$V_4 = V_g \frac{R_4}{R_4+R_5+R_6}$$ $$V_5 = V_g \frac{R_5}{R_4+R_5+R_6}$$

Since the voltmeter is an ideal high resistance voltmeter, now you have only to take the difference of the voltage at the two end of the voltmeter:

$$\Delta V_v = (V_4+V_5)-V_1 = V_g \left( \frac{R_4}{R_4+R_5+R_6} +\frac{R_5}{R_4+R_5+R_6}- \frac{R_1}{R_1+R_2+R_3}\right) $$

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  • $\begingroup$ Note that the sign of the reading on the voltmeter will depend on which way round the voltmeter is connected which is not specified in the diagram. $\endgroup$ – Peter Green Aug 3 '18 at 16:48

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