Question

A cell of e.m.f E and Internal Resistance r, r connected in series with a resistor and ammeter, A current of 0.8A is Observed to pass when the resistor is 2ohm when another resistor of 5 ohms is connected in parallel with the 2ohms resistor, the new ammeter reading is 1.0A.

a. Draw circuit diagram to illustrate the two arragement
b. using circuit diagram drawn in a above write down the equation for e.m.f E of the cell in each case.
c. calculate the internal resistance and e.m.f of the cell

Answer 1

Answer to Question #72561, Physics / Electric Circuits

A cell of e.m.f $E$ and Internal Resistance $r$ , connected in series with a resistor and ammeter. A current of 0.8A is observed to pass when the resistor is 2ohm. When another resistor of 5 ohms is connected in parallel with the 2ohms resistor, the new ammeter reading is 1.0A.

a. Draw circuit diagram to illustrate the two arrangement

b. using circuit diagram drawn in a above write down the equation for e.m.f $E$ of the cell in each case.

c. calculate the internal resistance and e.m.f of the cell

Solution.

a)

b)

For the first case:

$$E = I _ {1} \left(R _ {1} + r\right)$$

where resistance $R_{1} = 2$ ohms; current $I_{1} = 0.8A$

Answer to Question #72561, Physics / Electric Circuits

For the second case:

$$\frac {1}{R} = \frac {1}{R _ {1}} + \frac {1}{R _ {2}}$$ $$R = \frac {R _ {1} R _ {2}}{R _ {1} + R _ {2}}$$ $$E = I (R + r) = I \left(\frac {R _ {1} R _ {2}}{R _ {1} + R _ {2}} + r\right)$$

where $R$ is the total resistance of the circuit; $R_{2} = 5 \, ohms$; current $I = 1.0 \, A$

c)

For the first case:

$$E = 0.8 \cdot (2 + r)$$

For the second case:

$$E = 1 \cdot \left(\frac {2 \cdot 5}{2 + 5} + r\right) = \frac {10}{7} + r$$

Then:

$$0.8 \cdot (2 + r) = \frac {10}{7} + r$$ $$0.2 r = 1.6 - \frac {10}{7}$$ $$\frac {r}{5} = \frac {8}{5} - \frac {10}{7} = \frac {6}{35}$$

Answer:

$$r = 5 \cdot \frac {6}{35} = \frac {6}{7} \, ohms$$ $$E = 0.8 \cdot \left(2 + \frac {6}{7}\right) = \frac {4}{5} \cdot \frac {20}{7} = \frac {16}{7} \, V$$