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#1
Could any one please tell me how you would solve this question?

A battery is connected across a variable resistor. With the resistance set to 21 Ω, the current through the resistor is 0.48 A. When the resistor is at 36 Ω, the current is 0.30 A. What are the e.m.f. (E) and the internal resistance (r) of the battery?

r= 4 ohms
0
7 years ago
#2
(Original post by unknown123)
Could any one please tell me how you would solve this question?

A battery is connected across a variable resistor. With the resistance set to 21 Ω, the current through the resistor is 0.48 A. When the resistor is at 36 Ω, the current is 0.30 A. What are the e.m.f. (E) and the internal resistance (r) of the battery?

r= 4 ohms
Do you have any ideas on this yourself? What are you stuck on?

For a start, do you know a formula relating the emf of a battery to the internal resistance, the current in the circuit, and the resistance of the rest of the circuit?
You are given two sets of information and have two unknowns.
You need to solve this then as two simultaneous equations.
0
1 year ago
#3
(Original post by unknown123)
Could any one please tell me how you would solve this question?

A battery is connected across a variable resistor. With the resistance set to 21 Ω, the current through the resistor is 0.48 A. When the resistor is at 36 Ω, the current is 0.30 A. What are the e.m.f. (E) and the internal resistance (r) of the battery?

r= 4 ohms
can you show the calculation please...
0
10 months ago
#4
(Original post by nrnafisah)
can you show the calculation please...
I hope that you asked the question in a new thread instead of piggybag on an old thread in future. I would show you the calculation.

You can set up 2 equations and solve them using the following relationship.

E ‒ Ir = IRVar

where E is the emf, r is the internal resistance and RVar is the variable external resistance.

From “With the resistance set to 21 Ω, the current through the resistor is 0.48 A.”, we have

E ‒ 0.48r = 0.48 × 21

From “When the resistor is at 36 Ω, the current is 0.30 A.”, we have

E ‒ 0.30r = 0.30 × 36

We have two unknown emf E and internal resistance r and we have 2 independent equations to solve. Solving them gives emf E and internal resistance r.
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