Advertisements
Advertisements
Question
A cell of Emf 2 V and internal resistance 1.2 Ω is connected with an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in the diagram below:
1) What would be the reading on the Ammeter?
2) What is the potential difference across the terminals of the cell?
Advertisements
Solution 1
Given that
ε = 2 V, r = 1.2 Ω, RA = 0.8 Ω, R1 = 4.5 Ω, R2 = 9 Ω
1) We know that for the circuit
ε = IRtotal
Now, the total resistance of the circuit is
`R_"total" = r + R_A + R_p`
`1/R_p = 1/4.5 + 1/9 = 3/9`
∴ Rp = 3 Ω
⇒ Rtotal = 1.2 + 0.8 + 3 = 6 Ω
Hence, the current through the ammeter is
`I = epsilon/R_"total" = 2/6` = 0.33 A
2) The potential difference across the terminals of the cell is `V_"cell" = Ir = 0.33 xx 1.2` = 0.396 V
Solution 2
Given that
ε = 2 V, r = 1.2 Ω, RA = 0.8 Ω, R1 = 4.5 Ω, R2 = 9 Ω
1) We know that for the circuit
ε = IRtotal
Now, the total resistance of the circuit is
`R_"total" = r + R_A + R_p`
`1/R_p = 1/4.5 + 1/9 = 3/9`
∴ Rp = 3 Ω
⇒ Rtotal = 1.2 + 0.8 + 3 = 5 Ω
Hence, the current through the ammeter is
`I = epsilon/R_"total" = 2/5` = 0.4 A
2) The potential difference across the terminals of the cell is Vcell = Ecell - Ir
= `2 - (0.4 xx 1.2) = 2 - 0.48 V`
= 1.52 V
APPEARS IN
RELATED QUESTIONS
What happens to the resistivity of semiconductor with the increase of temperature?
A cell of e.m.f ε and internal resistance r is used to send current to an external resistance R. Write expressions for
- the total resistance of circuit.
- the current drawn from the cell.
- the p.d. across the cell.
- voltage drop inside the cell.
Explain why the p.d across the terminals of a cell is more in an open circuit and reduced in a closed circuit.
A cell is used to send current to an external circuit.
- How does the voltage across its terminals compare with its e.m.f.?
- Under what condition is the e.m.f. of a cell equal to its terminal voltage?
A cell of e.m.f. 1.8V and internal resistance 2Ω is connected in series with an ammeter of resistance 0.7Ω and a resistor of 4.5Ω as shown in Fig.
- What would be the reading of the ammeter?
- What is the potential difference across the terminals of the cell?
What is the colour code for the insulation on the earth wire?
A battery of 4 cell, each of e.m.f. 1.5 volt and internal resistance 0.5 Ω is connected to three resistances as shown in the figure. Calculate:
(i) The total resistance of the circuit.
(ii) The current through the cell.
(iii) The current through each resistance.
(iv) The p.d. across each resistance.
When a resistance of 3Ω is connected across a cell, the current flowing is 0.5 A. On changing the resistance to 7Ω, the current becomes 0.25A. Calculate the e.m.f. and the internal resistance of the cell.
The diagram in Figure shows a cell of e.m.f. ε = 4 volt and internal resistance r = 2 ohm connected to an external resistance R = 8 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when
- the key K is open, and
- the key K is closed
A battery of e.m.f. 6.0 V supplies current through a circuit in which the resistance can be changed. A high resistance voltmeter is connected across the battery. When the current is 3 A, the voltmeter reads 5.4 V. Find the internal resistance of the battery.
