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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?
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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
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