Question

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Q) A cell of emf 2v and interanal resistance 1.2 ohm and is connected to an ammeter of resistance 0.8 ohm and two resistors of 4.5 ohm and 9 ohm as shown in figure below

Find :-
a. the reading of the ammeter,

b.the potential difference accross the terminals of the cell , and

c.the potential difference accross the 4.5 ohm resistor m
​

Answer 1

Answer:

Given that =2V, r = I.2Ω, RA = O.8Ω,R1 = 4.5Ω , R2=9Ω ] 1. We know that for the circuit = IR total Now, the total resistance of the circuit is R total = r + RA + Rp 1/Rp = 1/4.5 + 1/9 = 3/9 Rp = 3Ω R total = 1.2 + 0.8 + 3 = 5Ω Hence, the current through the ammeter is I = V/Rtotal = 2/5 = 0.4A (ii) I = (E - V)/r ⇒ 0.4 = (2 - V)/1.2 I = v/R total = 2/5 = 0.4 AB ⇒ 0.48 = 2 – V - I V = 2 – 4.8 – 1.52V :. Potential difference Vcell 1.52

Explanation:

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Answer 2

Answer:

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