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Draw a schematic diagram of an electric circuit consisting of a battery of five 2V cells, a 20Ω resistor, a 30Ω resistor, a plug key, a connected in series. Calculate the value of current flowing through the 20Ω resistor and the power consumed by the 30Ω resistor.

Last updated date: 13th Jun 2024
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Answer
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Hint: In this value of two resistances are given which are connected in the series with the battery, so we first find the current flowing through the circuit by using ohms law and then we will find the current through 20Ω resistor and by using this current we will find the power consumed by the 30Ω resistor.

Complete step by step answer:
Let resistor
${R_1} = 20\Omega$
${R_2} = 30\Omega$
Since five 2V cells are connected in series, hence they will add up, and the voltage will be
$V = 10V$
It is given that all the components are connected in series with a key in series with them, so we will draw the circuit

Now, we know that ohm's law is given by the formula
$V = IR - - (i)$
In this circuit we can see the two resistors are connected in series, and we know in a series connection the resistance adds up; hence the equivalent resistance in this circuit becomes
${R_{eq}} = {R_1} + {R_2} \\ = 20 + 30 \\ = 50\Omega \\$
Now by using equation (i), we can write
$I = \dfrac{V}{R}$
Hence by substituting the value of resistance and voltage, we get the total current flowing in the circuit
$I = \dfrac{{10}}{{50}} \\ = 0.2A \\$
Now since both the resistor is in series hence the current flowing through them will be same; hence the current flowing through the 20Ω resistor will be $0.2A$
${I_{20\Omega }} = {I_{30\Omega }} = 0.2A - - (ii)$
We know the power consumed by a resistor is given by the resistor is given by the formula
$P = {I^2}R - - (iii)$
So the power consumed by the 30Ω resistor having current $0.2A$flowing through it will be
$P = {I^2}R \\ = {\left( {0.2} \right)^2} \times 30 \\ = \dfrac{2}{{10}} \times \dfrac{2}{{10}} \times 30 \\ = 1.2W \\$
Hence the power consumed by the 30Ω resistor $= 1.2W$

Note:Students must know that when cell are connected in series, then the total voltage increases and when they are connected in parallel then their capacity increases in ampere-hour.