Voltage in a parallel circuit [duplicate]

1. Why is voltage same across a parallel circuit? I mean what makes the voltage remain same across two resistors connected in parallel?

2. If an electric heater is connected in parallel with a bulb and if the bulb is then switched off, will the power dissipated in the heater increase?

We usually assume that wires are perfect conductors. In that case, the potential is the same at both ends of a wire, so at corresponding terminals of a parallel circuit. This is an approximation, valid when the resistors in the circuit are large compared to the wire resistance. For your last question, if the source is an ideal voltage source, switching te bulb off does not change the power in the heater.

Recall the definition of voltage: $~V= \frac{Work~done}{Unit~charge}$ Though current gets distributed, work done per charge remains the same. This is true only if, as Ross pointed out, $R_{wire}=0$, If it is not, then voltage across the resistor will be less than that across the source, because wire sections will start acting like individual resistors, and there'll be a voltage drop created across them.

NO, Power dissipated in heater will remain the same because as when connected in parallel, power dissipated by heater is = $P=I^{2}R~$ or $P=\frac{V^2}{R}$ (both are equal), and after the bulb is removed, $P$ remains equal to $I^{2}R~$ or $\frac{V^2}{R}$. However, total power dissipated will decrease when the bulb is removed. BTW Power is the rate at which energy is dissipated, or work is done. Power isn't dissipated. Energy is.

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• The answer to question 2 is not always true - you've made the additional assumption that the voltage across is constant with respect to load. – Alfred Centauri Oct 26 '14 at 11:37
• Internal Resistance? I almost never consider it while calcultaion. It haas a puny little effect on the answer. – user49111 Oct 26 '14 at 11:45
• I think that the voltage drop across the heater increases on switching off the bulb. beacause, initially when both were on,the current was splitting but when the bulb is switched off,the current no longer splits. the whole current goes to the heater only so V=IR, increase in I causes increase in V. and hence the power increses! but in reality that is not the case! why? – shraddha jain Oct 27 '14 at 10:52
• Voltage drop won't decrease! And therefore current flowing through heater will remain the same. Unless you consider internal resistance of the source, which will cause voltage to increase when bulb is removed, as Alfred said. – user49111 Oct 27 '14 at 11:15
• @shraddhajain, one cannot assume that the current that was through the bulb will now go through the heater unless the source is an ideal current source. In general, the 'whole' current will be different when the bulb is off than when it is on. In particular, if the source is an (effectively) ideal voltage source, the voltage across the heater will not (cannot) change at all when the bulb is switched off. Thus, the 'whole' current is reduced precisely by the amount of the bulb current before it was switched off. – Alfred Centauri Oct 28 '14 at 14:28

I mean what makes the voltage remain same across two resistors connected in parallel?

To be clear, in the context of ideal circuit elements, two resistors are parallel connected if all of the voltage across one resistor is across the other resistor.

This defines the parallel connection which is the dual of the series connection.

If the voltages across the resistors are not identical, they are not parallel connected.

Now, as others have pointed out, two resistors can be apparently parallel connected but, due to resistance in the connecting wires, one resistor will have less voltage across than the other.

However, it should be noted that the resistance of the wires are then in series with one of the resistors so the two resistors are not actually parallel connected.

That is to say, if one were to model and solve this circuit with ideal circuit elements, the connecting wires would be appear as resistors in the circuit diagram and then it would be plain to see that there is no parallel connection.

If an electric heater is connected in parallel with a bulb and if the bulb is then switched off, will the power dissipated in the heater increase?

It is possible and even likely that this will be the case. If the source providing the voltage to the parallel connected heater and bulb has internal resistance, then turning off the bulb will cause the voltage across the heater to increase.

Here's the relevant equation for when the bulb is on:

$$V_{heater/bulb} = V_{source} \frac{R_{heater}||R_{bulb}}{R_{source} + R_{heater}||R_{bulb}}$$

where $V_{source}$ is the open circuit source voltage and $R_{source}$ is the internal resistance of the voltage source.

When the bulb is switched off (or out of the circuit), the voltage increases to

$$V_{heater/bulb} = V_{source} \frac{R_{heater}}{R_{source} + R_{heater}}$$

The difference may be small but there will be a difference for non-zero $R_{source}$

Assuming the heater is simply a resistive heating element, the increased voltage across will increase the power delivered to the heater since

$$p_R = \frac{V^2_R}{R}$$

• +1. It said avoid commenting "+1", but I really have nothing else to say. – user49111 Oct 26 '14 at 11:47