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
Name two factors on which the internal resistance of a cell depends and state how does it depend on the factors stated by you.
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 battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm is series Find:
(a) the current through the battery
(b) the p.d. between the terminals of the battery.
A cell of emf. 1.5 V and internal resistance 10 ohms is connected to a resistor of 5 ohms, with an ammeter in series see fig.. What is the reading of the ammeter?
A cell supplies a current of 0.6 A through a 2Ω coil and a current of 0.3 A through on 8Ω coil. Calculate the e.m.f and internal resistance of the cell.
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.
(a) Calculate the total resistance across AB.
(b) If a cell of e.m.f 2.4 V with negligible internal resistance is connected across AB then calculate the current drawn from the cell.
Study the diagram:
- Calculate the total resistance of the circuit.
- Calculate the current drawn from the cell.
- State whether the current through 10 Ω resistor is greater than, less than or equal to the current through the 12 Ω resistor.
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
