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ICSE Class 10 Physics Question 19 of 20

Solved 2016 Question Paper ICSE Class 10 Physics — Question 15

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Question 9(c)

A battery of e.m.f. 12 V and internal resistance 2 Ω is connected with two resistors A and B of resistance 4 Ω and 6 Ω respectively joined in series. [4]

A battery of e.m.f. 12 V and internal resistance 2 Ω is connected with two resistors A and B of resistance 4 Ω and 6 Ω respectively joined in series. Find (i) Current in the circuit (ii) The terminal voltage of the cell (iii) The potential difference across 6 Ω Resistor (iv) Electrical energy spent per minute in 4 Ω Resistor. ICSE 2016 Physics Solved Question Paper.

Find:

(i) Current in the circuit.

(ii) The terminal voltage of the cell.

(iii) The potential difference across 6 Ω Resistor.

(iv) Electrical energy spent per minute in 4 Ω Resistor.

Answer

Given,

RA = 4 Ω

RB = 6 Ω

e.m.f. = 12 V

r = 2 Ω

(i) Two resistors ( 4 Ω and 6 Ω) are connected in series, so Rs = 4 + 6 = 10 Ω

From relation,

I = 𝐸𝑅+𝑟\dfrac{𝐸}{𝑅 + 𝑟}

Substituting the values we get,

I=1210+2I=1212⇒I=1AI = \dfrac{12}{10 + 2} \\[0.5em] I = \dfrac{12}{12} \\[0.5em] \Rightarrow I = 1 A

(ii) Terminal voltage of cell V = ε - Ir

Substituting the values we get,

V=12−(1×2)V=12−2⇒V=10VV = 12 - (1 \times 2) \\[0.5em] V = 12 - 2 \\[0.5em] \Rightarrow V = 10 V

(iii) P.d. across 6 Ω resistor = I R6Ω

Substituting the values we get,

V6Ω = 1 x 6 = 6 V

(iv) Electrical energy spent per minute (60 s) in 4 Ω Resistor = I2RAt = 1 x 6 = 6 V