Lesson 2, Topic 5
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Electromotive Force (E.M.F)

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E.m.f in Series:

When cells are connected in series, the positive terminal of one is connected to the negative terminals of the other. With this arrangement, a larger emf is obtained in the circuit with improved current.

Etotal = E1 + E2 + E3

rtotal = r1 + r2 + r3

Where r = the internal resistance of the cell.

E.m.f in Parallel:

For a parallel connection, all positive terminals are joined together and all negative terminals are joined together.

The e.m.f of one of the cell is equal to the total e.m.f E1 = E2 = E3

Example 1:

A cell of e.m.f 1.5V is connected to a resistor of resistance 3Î©. Calculate the current flowing.

Solution:

Emf = 1.5V, R= 3Î©, I =?

V = IR

I = $$\frac{V}{R}$$

I = $$\frac{1.5 V}{3 \Omega}$$

I = 0.5A

Example 2:

A battery of emf 24v and internal resistance 4Î© is connected to a resistor of 32Î©. What is the terminal p.d of the battery?

Solution:

Whether you are asked to or not, always sketch a circuit diagram before solving any question on current electricity.

First, calculate the current, I

I = $$\frac {E}{R + r}$$

= $$\frac {24}{32 + 4 }$$

= $$\frac {24}{36 }$$

= $$\frac {2}{3} \scriptsize A$$

or 0.67A

âˆ´ Terminal p.d, V = IR

= 0.67 x 32

= 21.3v

Evaluation Question:

1. A cell of e.m.f 1.5V and internal resistance 2.5Î© is connected in series with an ammeter of resistance 0.5Î© and a load of resistance 7.0Î©. Calculate the current in the circuit.
2. A cell of e.m.f 1.5V is connected in series with a resistor of resistance 3.0Î©. A voltmeter connected across the cell registers 0.9V, calculate the internal resistance of the cell.

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