help! two quick Dc electricity questions

Three identical resistors are connected across a potential difference V so that one of them is in parallel with the other two which are connected in series. The power dissipated through the first one, compared to the power dissipated by each of the other two, is approximately
A the same

B half as much

C twice as much

D four times as much

Firstly, how do you determine the current that flows through a branch? Like if I wanted to know the current that flows through the branch with two resistors, would it be V/2R?

Lastly, can someone explain in detail how the potential difference and current in a circuit can be varied using a rheostat? (please explain in terms of V=IR for the rheostat and the component in the circuit)


thanks!

A diagram always helps.
Yes. So the current in the branch with one resistor is
V/R
and the current in the branch with the 2 resistors in series is
V/2R
The current in the single resistor is twice that in the other branch.
The pd is the same for both branches as they are in parallel.
So the power in the single resistor is I²R
In the other branch the current is half, so in either of the other two resistors, half the current will give (½)² the power. That is 4 times less.
So it's answer D




A rheostat is a variable resistor, so if you vary the resistance in a circuit you can vary the current because I = V/R
Increase R and you reduce I.

In some other part of the circuit, if you have a component with resistance R and you change the current in it (using the rheostat as explained before) you change the voltage across it because V=IR
Increasing I will increase V

For the variation of current with the rheostat, the voltage available in the circuit was constant. But for the variation of voltage with the rheostat, was the resistance of the component constant? Why is this so? Shouldn't the new current flowing through the component be constant since the components in series?

In this diagram, Bulb A has a voltage of 12V and the battery has a voltage of 15V. Since the lines in branches have the same voltage,would this mean that the sum of the voltages of bulb B and R2 will equal 12V?
Also, between which two points is the voltmeter measuring the voltage in this diagram;which components?

Again in this diagram, imagine R2 was a variable resistor and that its resistance has been decreased. By using V=IR, could you explain how the voltage,current and resistance for R1 and bulb B will change?

Using the rheostat to vary current assumed that the source voltage was constant. So I=V/R_t where R_t is the resistance of the whole circuit. This resistance is increased by increasing the resistance of the rheostat which is placed in the circuit.
Varying the voltage across a component that has a fixed resistance can be achieved by varying the current through it. So if you change the current in a circuit, you change the voltages.
The problem is that the effect of the rheostat will depend on exactly how it has been connected to the circuit. It's impossible to answer the question specifically without being given a specific circuit. This links to your last question.


1. Yes that's correct.

2. It's measuring the voltage across bulb A (12V) and sum of the voltages across bulb B and R2 (also 12V)


If you decrease R2 you will increase the current in that branch so bulb B gets more current. By decreasing R2 you also decrease the total resistance of the whole circuit so this will increase the overall current through the cell and through R1. The value of R1 will not change as it's a fixed resistor.

With circuits like this the only way to find out exactly what happens to the values of currents and voltages is to apply Ohm's Law to the individual components. The result is an interaction between all the components.

Thanks! thats cleared a lot for me

One last thing, how can you tell the voltmeter only gives the voltage of the branches in front of it; the voltmeter is also connected across the battery which is at 15V...
Is it because that, even though its connected across the same two points, the battery is not a part of the junction?

Three identical resistors are connected across a potential difference V so that one of them is in parallel with the other two which are connected in series. The power dissipated through the first one, compared to the power dissipated by each of the other two, is approximately
A the same
B half as much
C twice as much
D four times as much
Firstly, how do you determine the current that flows through a branch? Like if I wanted to know the current that flows through the branch with two resistors, would it be V/2R?
Lastly, can someone explain in detail how the potential difference and current in a circuit can be varied using a rheostat? (please explain in terms of V=IR for the rheostat and the component in the circuit)
thanks!