Question

1. Two cells, each of e.mf. 1.5 V and internal resistance 1ohm are connected in parallel, to form a battery. The
battery is connected to an external resistance of 0.5 ohm and two resistances of 3 ohm and 1.5 ohm in parallel.
(a) Draw the circuit diagram .
(b) Calculate the current in the
main circuit.
(c) Calculate the current in 1.5 ohm resistor .
(d) Calculate the drop in potential across the terminal of the battery.
Ans = ((b) 0.5 A , (c) 0.25 A , (d) 0.375 V).
Help please
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Answer 1

$\tt{ \underline{✪ \: Qᴜᴇsᴛɪᴏɴ:-}}$

→ Two cells, each of e.mf. 1.5 V and internal resistance 1ohm are connected in parallel, to form a battery. The

battery is connected to an external resistance of 0.5 ohm and two resistances of 3 ohm and 1.5 ohm in parallel.

(a) Draw the circuit diagram .

(b) Calculate the current in the

main circuit.

(c) Calculate the current in 1.5 ohm resistor .

(d) Calculate the drop in potential across the terminal of the battery.

Ans = ((b) 0.5 A , (c) 0.25 A , (d) 0.375 V).

$\tt{ \underline{✪ ᴀɴsᴡᴇʀ:-}}$

$\tt{✪ \: ᴀɴsᴡᴇʀ \: (ʙ) : }$

→ Current in the main circuit

→ Total internal resistance of two parallel cells

→ I/R1 = 1 + 1

→ R = 1/2 = 0.5Ω

→ Effective resistance between

→ PQ = 1/R2 = 1/3 + 2/3 R2 = 1Ω

→ Total resistance of circuit R = R1 + R2+ external resistance

→ R = 0.5 + I + 0.5 = 2Ω, V= 1.5V current in main circuit I = V/R = 1.5/2 = 0.75A

$\tt{ \underline{✪ \: ᴀɴsᴡᴇʀ \: (ᴄ) \: ✪}}$

→ Current in 1 .5Ω resistor

→ P.d. between PQ = IR^2

→ V1 = 0.75 × 1

→ V1 = 0.75V

→ V1 = 0.75

→ I1= V1 /1.5 = 0.75/1.5 = 0.5A

$\tt{ \underline{✪ \: ᴀɴsᴡᴇʀ \: (ᴅ) \: ✪ }}$

→ Drop in potential across the terminals of battery E – V = Ir = 0.75 × 0.5 = 0.375V