a resistance of 4 ohm and a wire of length 5 metres and resistance 5 ohm are joined in series and connected to a cell of EMF 10 volt and resistance 1 ohm The parallel combination of two identical cells is balanced across 300 centimetre of the wire the EMF of a cell is
Answers
Answer:
The answer will be 3 V
Explanation:
According to the problem the wire with 2 resistor of 4 ohm and 5 ohm and the length of 5 m are connected in series with a cell having 10 volt emf and this combination is in parallel combination with two identical cells.
Therefore the total current in the wire, 10/ 5+4+1 = 1 A [ by using the formula of V = IR]
Therefore for 300 centimeter of wire the voltage will be 3 volt
As the cells are identical therefore it will have the same emf of 3V each.
Answer:3V
Explanation:
According to the problem the wire with 2 resistor of 4 ohm and 5 ohm and the length of 5 m are connected in series with a cell having 10 volt emf and this combination is in parallel combination with two identical cells.
Therefore the total current in the wire, 10/ 5+4+1 = 1 A [ by using the formula of V = IR]
Therefore for 300 centimeter of wire the voltage will be 3 volt
As the cells are identical therefore it will have the same emf of 3V each.
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