Resistance help

Hi!

I'm really struggling with the following AS Level question so any help or guidance would be really appreciated.

A high resistance voltmeter is connected across a cell and the reading is 1.6 V. When a 3.0 Ω resistor is put across the cell the reading falls to 1.5 V. Why does it fall, and what is the internal resistance of the cell?

Thank you.

Cathy

Reply 1

Bape

2

CathyLou

Hi!

I'm really struggling with the following AS Level question so any help or guidance would be really appreciated.

A high resistance voltmeter is connected across a cell and the reading is 1.6 V. When a 3.0 Ω resistor is put across the cell the reading falls to 1.5 V. Why does it fall, and what is the internal resistance of the cell?
Thank you.
Cathy

E = I(R+r) E = IR +Ir

V = IR so E = V + Ir

E=emf, V=terminal pd, Ir=lost volts

1.6V = 1.5V + Ir

Ir = 0.1V

3ohm resistor has 1.5V across too since parallel with voltmeter
I=V/R I=1.5V/3ohm I=0.5A

so 0.5A x r = 0.1V r = 0.2ohms

I havent done internal resistance for like a year now, but i think thats right

Reply 2

Morbo

17

When the high resistance voltmeter is put across the cell, virtually no current flows through the cell. When the resistor is put across the cell, some current is flowing through the cell, which has resistance.

Reply 3

CathyLou

OP

2

Thanks to both of you for your help!

Cathy

Reply 4

CathyLou

OP

2

I'm also really stuck on this question too, and have no idea how to even start it, so any help would be really appreciated.

A potential divider circuit is set up with a LDR and a fixed resistor, R. The LDR has a resistance of 200 Ω when it is light and 2500 Ω when it is dark. A motor is set so that it comes on if 4.0 V is applied across it. A 6.0 V battery provides the emf (assume it has no internal resistance).

When it is dark, does the motor come on if R is (a) 3000 Ω (b) 4000 Ω (c) 5000 Ω?

When it is light, does the motor come on if R is (a) 300 Ω (b) 400 Ω (c) 500 Ω?

If the motor were moved so that it is across the LDR instead, what would R have to be so that the motor came on in the light?

Thank you.

Cathy

Reply 5

Theorist

10

CathyLou

I'm also really stuck on this question too, and have no idea how to even start it, so any help would be really appreciated.

A potential divider circuit is set up with a LDR and a fixed resistor, R. The LDR has a resistance of 200 Ω when it is light and 2500 Ω when it is dark. A motor is set so that it comes on if 4.0 V is applied across it. A 6.0 V battery provides the emf (assume it has no internal resistance).

When it is dark, does the motor come on if R is (a) 3000 Ω (b) 4000 Ω (c) 5000 Ω?

When it is light, does the motor come on if R is (a) 300 Ω (b) 400 Ω (c) 500 Ω?

If the motor were moved so that it is across the LDR instead, what would R have to be so that the motor came on in the light?

From the last statement of the question, I'm assuming initially that the motor is connected in parallel across the resistor, and that the motor is of an infinite resistance so it does not change the effective resistance of the resistor-motor combination.

If this is the case, use the fact that in a potential divider where both the components are in series with each other,

I

is common and therefore

V\propto R

.

When it is dark, if the motor is to work, the potential difference across the LDR must be

2\,\mathrm{V}

which is

\frac{\epsilon}{3}

. As potential difference is directly proportional to voltage, the resistance therefore the LDR must be

\frac{1}{3}

of the total circuit resistance. Therefore, the resistance of the resistor must be double that of the LDR, so it is

5000\,\Omega

.

That's how you do the first part. The other parts require the same method, you just have to alter the figures and ensure that you know which components will take more voltage if the motor is to work.

Reply 6

CathyLou

OP

2

Thanks so much for your help!

Cathy

Resistance help

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