
Suppose the resistors ${R_1}$, ${R_2}$ and ${R_3}$ have the values $5\Omega $,$10\Omega $ and $30\Omega $ respectively. Which have been connected to a battery of $12V$ in a parallel connection. Calculate
1. Current through each resistor
2. Total current in the circuit
3. Total circuit resistance
Answer
559.5k+ views
Hint:Current through each resistor can be calculated applying Ohm’s law as we are provided with voltage of the circuit and the individual resistances of each resistor.Total current can be calculated using the Ohm’s law as the voltage of the circuit is already given.Resistances are given in the parallel connection.
Complete Step by Step Answer:
GIVEN THAT ${R_1} = 5\Omega $ ${R_2} = 10\Omega $ ${R_3} = 30\Omega $ $V = 12Volt$
1. Firstly, calculating current through each resistor:
For ${R_1} = 5\Omega $
Using ohm’s law ${I_1} = \dfrac{V}{{{R_1}}}$
${I_1} = \dfrac{{12}}{5} = 2.4A$
Like ways we can calculate current through each resistor.
For ${R_2} = 10\Omega $
${I_2} = \dfrac{V}{{{R_2}}} = \dfrac{{12}}{{10}}$
$\Rightarrow{I_2} = 1.2A$
For ${R_3} = 30\Omega $
$\Rightarrow{I_3} = \dfrac{V}{{{R_3}}} \\
\Rightarrow{I_3} = \dfrac{{12}}{{30}}$
$\therefore{I_3} = 0.4A$
2. Total current in the circuit$\left( I \right)$:
Firstly, do part $(3)$solved below
Using Ohm’s law
$V = IR$
$\Rightarrow I = \dfrac{{12}}{3} \\
\therefore I = 4A$
TOTAL CURRENT IS $4A$.
3. Total circuit resistance $\left( R \right)$: As we all know from the question the resistances are connected in parallel, applying the resistances formula
$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}$
$\Rightarrow\dfrac{1}{R} = \dfrac{1}{5} + \dfrac{1}{{10}} + \dfrac{1}{{30}}$
$\Rightarrow\dfrac{1}{R} = \dfrac{{6 + 3 + 1}}{{30}} = \dfrac{{10}}{{30}}$
$\therefore R = 3\Omega $
TOTAL RESISTANCE IS $3\Omega $.
Note:Resistors are said to be connected in parallel when both of their terminals are respectively connected to each terminal of the other resistor or resistors. The current in each branch of a parallel connection circuit is different. Resistors in a parallel have common voltage across them.
Complete Step by Step Answer:
GIVEN THAT ${R_1} = 5\Omega $ ${R_2} = 10\Omega $ ${R_3} = 30\Omega $ $V = 12Volt$
1. Firstly, calculating current through each resistor:
For ${R_1} = 5\Omega $
Using ohm’s law ${I_1} = \dfrac{V}{{{R_1}}}$
${I_1} = \dfrac{{12}}{5} = 2.4A$
Like ways we can calculate current through each resistor.
For ${R_2} = 10\Omega $
${I_2} = \dfrac{V}{{{R_2}}} = \dfrac{{12}}{{10}}$
$\Rightarrow{I_2} = 1.2A$
For ${R_3} = 30\Omega $
$\Rightarrow{I_3} = \dfrac{V}{{{R_3}}} \\
\Rightarrow{I_3} = \dfrac{{12}}{{30}}$
$\therefore{I_3} = 0.4A$
2. Total current in the circuit$\left( I \right)$:
Firstly, do part $(3)$solved below
Using Ohm’s law
$V = IR$
$\Rightarrow I = \dfrac{{12}}{3} \\
\therefore I = 4A$
TOTAL CURRENT IS $4A$.
3. Total circuit resistance $\left( R \right)$: As we all know from the question the resistances are connected in parallel, applying the resistances formula
$\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}$
$\Rightarrow\dfrac{1}{R} = \dfrac{1}{5} + \dfrac{1}{{10}} + \dfrac{1}{{30}}$
$\Rightarrow\dfrac{1}{R} = \dfrac{{6 + 3 + 1}}{{30}} = \dfrac{{10}}{{30}}$
$\therefore R = 3\Omega $
TOTAL RESISTANCE IS $3\Omega $.
Note:Resistors are said to be connected in parallel when both of their terminals are respectively connected to each terminal of the other resistor or resistors. The current in each branch of a parallel connection circuit is different. Resistors in a parallel have common voltage across them.
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