Need help with a question on potential difference

[StarlitVarsh]

Homework Statement

Consider the circuit shown in the figure. Calculate the potential difference across 3 Ohm resistor

Relevant Equations

I believe V=IR and, R = R1+R2+ R3, 1/R = 1/R1 + 1/R2 + 1/R3... are relevant.

So far, here's what I have:
Taking the parallel-connected resistors alone, 1/Total effective resistance (R) = 1/3ohm + 1/6ohm = 1/2, so R = 2ohm

Replacing the 3 and 6 ohm resistors with a 2 ohm resistor, the total resistance will be 2+4 = 6ohm

Using V=IR,
I (current) = V(potential difference)/R(resistance)
=12V/6ohm = 2A

I've found the total current, but now I'm pretty much stuck???

 

[haruspex]

Homework Statement: Consider the circuit shown in the figure. Calculate the potential difference across 3 Ohm resistor
Relevant Equations: I believe V=IR and, R = R1+R2+ R3, 1/R = 1/R1 + 1/R2 + 1/R3... are relevant.

View attachment 332778

So far, here's what I have:
Taking the parallel-connected resistors alone, 1/Total effective resistance (R) = 1/3ohm + 1/6ohm = 1/2, so R = 2ohm

Replacing the 3 and 6 ohm resistors with a 2 ohm resistor, the total resistance will be 2+4 = 6ohm

Using V=IR,
I (current) = V(potential difference)/R(resistance)
=12V/6ohm = 2A

I've found the total current, but now I'm pretty much stuck???

What is the p.d. across the 4Ω?

 

[StarlitVarsh]

What is the p.d. across the 4Ω?

Honestly, I'm not sure. I learnt the concept just a couple hours ago, so I'm not sure how most of the questions in the topic work.

 

[DaveE]

The potential difference (voltage) across a resistor is the current through it times the resistance; V=I⋅R.
Current flows from the positive voltage side to the negative side, except in voltage sources, like batteries, where it's the opposite. In any complete loop of the circuit all of the voltages must sum to zero; but be mindful of the voltage polarities when you do that sum.

 

[Steve4Physics]

Using V=IR,
I (current) = V(potential difference)/R(resistance)
=12V/6ohm = 2A

I've found the total current, but now I'm pretty much stuck???

That's good so far. The 'total current' is 2A. That means the current through the cell is 2A. And the current through the 4Ω resistor is therefore also 2A (since the cell and resistor are in series, they have the same current).

Remember, you can use 'V=IR' more than once. You can use it for any resistor providing that: V is the p.d. across that resistor; I is the current through that resistor; and R is the resistance of that resistor.

There is more than one way to do the problem. But, since you've already got the total current, one method is to find the p.d. across the 4Ω resistor and see if you can then use this value.

 

[StarlitVarsh]

The potential difference (voltage) across a resistor is the current through it times the resistance; V=I⋅R.
Current flows from the positive voltage side to the negative side, except in voltage sources, like batteries, where it's the opposite. In any complete loop of the circuit all of the voltages must sum to zero; but be mindful of the voltage polarities when you do that sum.

I found the solution soon after my initial post, but thanks anyways!

 

[StarlitVarsh]

That's good so far. The 'total current' is 2A. That means the current through the cell is 2A. And the current through the 4Ω resistor is therefore also 2A (since the cell and resistor are in series, they have the same current).

Remember, you can use 'V=IR' more than once. You can use it for any resistor providing that: V is the p.d. across that resistor; I is the current through that resistor; and R is the resistance of that resistor.

There is more than one way to do the problem. But, since you've already got the total current, one method is to find the p.d. across the 4Ω resistor and see if you can then use this value.

I found the solution by taking the total effective resistance of the 3 and 6-ohm resistors, and multiplying with the current of the entire circuit - and the solution I found was accurate to the answer key. Thanks for the suggestion, though!

 

[Steve4Physics]

I found the solution by taking the total effective resistance of the 3 and 6-ohm resistors, and multiplying with the current of the entire circuit - and the solution I found was accurate to the answer key. Thanks for the suggestion, though!

That's a good way to do it.

For information, here are another couple of methods. We note that the 3Ω and 6Ω resistors are in a parallel (total effective resistance =2Ω) and therefore have the same pd. Call this pd 'U'. We are being asked to find U.

1. The pd across the 4Ω resistor is V = IR = 2A x 4Ω = 8V.
Going round the circuit 'loop', the pds must add up to equal the cell's emf:
8+U = 12, giving U.

2. When you've learned about the potential divider formula, you can immediately write down U = 12 $\times \frac 2{4+2}$.