Three wires each of 30 ohm resistance, are arranged in parallel and connected to a battery
ef 10 lecianche cells, each having an EMF of 15 volts. The main current is
A) 0.1 amp
(B) 0.7 amp
(C) 0.375 amp
(D) 0.2 amp
Answers
Given:
Three wires each of 30-ohm resistance
The wires are arranged in parallel and connected to a battery of 10 Leclanche cells
Emf of each Lechlanche cell = 15 volts
To find:
The main current
Solution:
Finding the total emf:
Let's assume that we have 10 Leclanche cells, each with an emf of 15 volts (say E₁ = 15 V, E₂ = 15 V ...... ), connected in a series combination having the same polarity.
∴ Net Emf = E₁ + E₂ + ....... + E₁₀ = 15 + 15 + ..... (10 times) = 15 × 10 = 150 volts
Finding the equivalent resistance:
We have 3 wires each of 30 Ω (say R₁ = 30 Ω, R₂ = 30 Ω & R₃ = 30 Ω) connected in parallel combination, therefore, net resistance in parallel combination is given by,
Now, Finding the main current through the circuit:
By using Ohm's Law, we have
where
I = Current (in amperes)
V = Voltage (in volts)
R = Resistance (in ohms)
Here, we have
∴
Thus, the main current through the given circuit is 15 A.
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four cells, each of EMF 2 volt and internal resistance 2 ohm are connected in parallel to form battery, the battery is connected to external resistance of 2.5 ohm and two resistance of 6 ohm and 3 ohm in parallel draw a circuit diagram and answer what is current in the main circuit? what is current in 6 ohm wire? what is drop in potential?
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Answer:
Step-by-step explanation: