Question

Three wires each of 30 ohm resistance, are arranged in parallel and connected to a battery of 10 leclanche cells, each having an EMF of 15 volts. The main current is

Answer 1

Given:

3 wires each of 30 ohm resistance are arranged in parallel combinations and connected to 10 LecLanche cells each having an EMF of 15 volts.

To find:

Current through the resistances.

Concept:

We need to separately calculate the total EMF provided by the cells as well the total resistance of those wire.

Then we can apply Ohm's Law to get the value of current in the circuit.

Calculation:

Let's consider that the LecLanche cells are connected side to side i.e. in series combination with similar polarity.

Hence net EMF will be :

$(EMF)_{net} = 15 + 15 + ... 10 \: times$

$= > (EMF)_{net} = 10 \times 15$

$= > (EMF)_{net} = 150 \: volts$

The resistance wires are in parallel combination:

Net resistance be R_{eq}

$\therefore \dfrac{1}{R_{eq}} = \dfrac{1}{30} + \dfrac{1}{30} + \dfrac{1}{30}$

$= > \dfrac{1}{R_{eq}} = \dfrac{1 + 1 + 1}{30}$

$= > \dfrac{1}{R_{eq}} = \dfrac{3}{30}$

$= > \dfrac{1}{R_{eq}} = \dfrac{1}{10}$

$= > R_{eq} = 10 \: ohm$

So the net current through the circuit will be:

$= > current = \dfrac{(EMF)_{net}}{R_{eq}}$

$= > current = \dfrac{150}{10}$

$= > current = 15 \: amp$

So final answer:

Current through the circuit is 15 Ampere.