Question

A cell of e.m.f. E and internal resistance r was

connected in series with two series external resistors, A
(of 8-ohms) and B (of 2-ohms). A high resistance
voltmeter connected across A was found to read 8volts.
When another resistor C (of 8 ohms) was connected
parallel to A, and then across A and C, the voltmeter read
(i) Draw the circuit diagrams of the two arrangements.
(ii) Calculate the internal resistance of the cell. (iii)Calculate the e.m.f. of the cell.

Answer 1

(i) Circuit diagrams of the two arrangements:

Arrangement 1:


___ R_A _______ R_B
| |-------| |
-----| E | | |
|___|-------|_______|
Arrangement 2:


___ R_A _______ R_B
| |-------| |
-----| E | | R_C |
|___|-------|_______|
(ii) To calculate the internal resistance of the cell, we can use the formula for the potential difference across the external resistor A:

V_A = E - I * r

where V_A is the potential difference across A, E is the emf of the cell, I is the current passing through the circuit, and r is the internal resistance of the cell.

In the first arrangement, the potential difference across A is given as 8 volts. Since there is no parallel resistor connected, the current passing through the circuit is the same as the current passing through resistor A.

V_A = 8 volts
R_A = 8 ohms
I = V_A / R_A = 8 volts / 8 ohms = 1 ampere

Now we can calculate the internal resistance r:

V_A = E - I * r
8 volts = E - 1 ampere * r

Since the potential difference across A is equal to the emf of the cell in this arrangement, we have:

E - r = 8 volts ---- (Equation 1)

In the second arrangement, the potential difference across A changes when resistor C is connected in parallel with A. Let's say the new potential difference across A is V_A'.

When resistor C is connected in parallel, the total resistance in parallel is given by:

1 / R_total = 1 / R_A + 1 / R_C

1 / R_total = 1 / 8 ohms + 1 / 8 ohms
1 / R_total = 1 / 4 ohms
R_total = 4 ohms

Now we can calculate the current passing through the circuit with C connected in parallel:

I' = V_A' / R_total

Since the voltmeter reads 8 volts across A and C, the potential difference across A is still 8 volts, and thus V_A' = 8 volts.

I' = 8 volts / 4 ohms = 2 amperes

Using the same formula for potential difference across A, we have:

V_A' = E - I' * r

Substituting the known values:

8 volts = E - 2 amperes * r ---- (Equation 2)

To solve for the internal resistance r, we can subtract Equation 1 from Equation 2:

(E - 2r) - (E - r) = 8 volts - 8 volts
-E + 2r + Er = 0
3r = E
r = E / 3

(iii) To calculate the emf of the cell, we can substitute the value of r obtained in Equation 1:

E - r = 8 volts
E - (E / 3) = 8 volts
(3E - E) / 3 = 8 volts
2E / 3 = 8 volts
2E = 3 * 8 volts
2E = 24 volts
E = 12 volts

Therefore, the internal resistance of the cell is E / 3 or 12 volts / 3, which is 4 ohms. The emf of the cell is 12 volts

Answer 2

Step-by-step explanation:

(i) Circuit Diagrams:

Original circuit:

----[ Cell (E, r) ]----[ Resistor A (8 Ω) ]----[ Resistor B (2 Ω) ]----

After adding resistor C in parallel to A:

----[ Cell (E, r) ]----[ Resistor A (8 Ω) ]----[ Resistor B (2 Ω) ]----

|

[ Resistor C (8 Ω) ]

(ii) Calculation of internal resistance of the cell:

In the original circuit, the voltmeter reading across resistor A is 8 volts. This means that the potential difference across resistor A is equal to the voltage provided by the cell, E.

Using Ohm's Law, we can calculate the current passing through resistor A:

V = I * R

8 = I * 8

I = 1 A (Amperes)

The current passing through resistor A is 1 A. Since resistor B is in series with resistor A, the same current passes through resistor B as well.

Using the concept of potential dividers, we can find the potential difference across the internal resistance of the cell, r:

V_r = I * r

Since the voltmeter reading across resistor A is 8 volts, and resistor B has a resistance of 2 Ω, the potential difference across the internal resistance can be calculated:

V_r = V_A - V_B

V_r = 8 - (1 * 2)

V_r = 6 volts

Now we can calculate the internal resistance of the cell using Ohm's Law:

V_r = I * r

6 = 1 * r

r = 6 Ω

Therefore, the internal resistance of the cell is 6 Ω.

(iii) Calculation of the e.m.f. of the cell:

Using Ohm's Law, we can find the voltage provided by the cell, E:

E = V_A + (I * r)

E = 8 + (1 * 6)

E = 14 volts

Therefore, the e.m.f. of the cell is 14 volts.