https://postimg.cc/PPvPZRpX

Im not sure how ΔV2 = 4v, ΔV1 = 4v and how ΔV3 = ΔVbattery = 8v

Im also confused on how a voltmeter would be connected across a and b

Im not sure how ΔV2 = 4v, ΔV1 = 4v and how ΔV3 = ΔVbattery = 8v

Im also confused on how a voltmeter would be connected across a and b

Original post by MonoAno555

https://postimg.cc/PPvPZRpX

Im not sure how ΔV2 = 4v, ΔV1 = 4v and how ΔV3 = ΔVbattery = 8v

Im also confused on how a voltmeter would be connected across a and b

Im not sure how ΔV2 = 4v, ΔV1 = 4v and how ΔV3 = ΔVbattery = 8v

Im also confused on how a voltmeter would be connected across a and b

Assuming youre told the battery is 8v, then its just a circuit with two parallel branches, so R1+R2 and R3 and the potential difference across each branch is the same and equal to the battery. For the R1+R2 branch, the current flowing through them is the same, as are the resistances, so the potential difference across each is the same and sums to 8v.

A voltmeter simply connects to two points on a circuit. You can (ideally) model it as another branch but with infinitely high resistance so no current flows through it. However the potential difference will be the same as the potential difference between the two points in the circuit as thats what its measuring.

Forgot to upload part b of the actual question I'm stuck on its:

A student uses a voltmeter and reads a value of 4V when it is connected

across points “a” and “b”. Determine values of the voltmeter readings

across each device in the circuit, ∆V1, ∆V2, ∆V3 and ∆VBattery. Briefly

explain your reasoning.

A student uses a voltmeter and reads a value of 4V when it is connected

across points “a” and “b”. Determine values of the voltmeter readings

across each device in the circuit, ∆V1, ∆V2, ∆V3 and ∆VBattery. Briefly

explain your reasoning.

Original post by MonoAno555

Forgot to upload part b of the actual question I'm stuck on its:

A student uses a voltmeter and reads a value of 4V when it is connected

across points “a” and “b”. Determine values of the voltmeter readings

across each device in the circuit, ∆V1, ∆V2, ∆V3 and ∆VBattery. Briefly

explain your reasoning.

A student uses a voltmeter and reads a value of 4V when it is connected

across points “a” and “b”. Determine values of the voltmeter readings

across each device in the circuit, ∆V1, ∆V2, ∆V3 and ∆VBattery. Briefly

explain your reasoning.

Ok, the previous answer wasnt quite rght as youre told Delta V2, rather than the potential difference across the battery (otherwise the reasoning is ok). So as the pot diff from a to b is 4v (the potential difference across R2) you should be able to alter the previous explanation? Have a go and upload what you tried if youre still stuck.

Original post by mqb2766

Ok, the previous answer wasnt quite rght as youre told Delta V2, rather than the potential difference across the battery (otherwise the reasoning is ok). So as the pot diff from a to b is 4v (the potential difference across R2) you should be able to alter the previous explanation? Have a go and upload what you tried if youre still stuck.

Spent more time on it and think i got it:

4V across "ab" means V2 = 4v.

As R1 = R2, V1 = 4v via potential divider theory, as the equal resistance means R1 and R2 have equal voltage share in series.

Via Kirchoff's first law, there's 8v total for 1 full loop, hence for the first loop connected to the Battery R3 = VBAT = 8v for 1 resistor of R3 in that loop

Diagram with voltmeter I drew:

https://postimg.cc/JsfK0W7B

Original post by MonoAno555

Spent more time on it and think i got it:

4V across "ab" means V2 = 4v.

As R1 = R2, V1 = 4v via potential divider theory, as the equal resistance means R1 and R2 have equal voltage share in series.

Via Kirchoff's first law, there's 8v total for 1 full loop, hence for the first loop connected to the Battery R3 = VBAT = 8v for 1 resistor of R3 in that loop

Diagram with voltmeter I drew:

https://postimg.cc/JsfK0W7B

4V across "ab" means V2 = 4v.

As R1 = R2, V1 = 4v via potential divider theory, as the equal resistance means R1 and R2 have equal voltage share in series.

Via Kirchoff's first law, there's 8v total for 1 full loop, hence for the first loop connected to the Battery R3 = VBAT = 8v for 1 resistor of R3 in that loop

Diagram with voltmeter I drew:

https://postimg.cc/JsfK0W7B

Pretty much, Id have said that for R1 and R2, the current is the same (in series), as are the two resistances, so the voltage across both is the same. Similar to what you said, but mention the current is the same.

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