Three resistors $2\Omega$, A, B and C, are connected as shown in figure. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors.
Answer
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Hint: Use the formula for power dissipated by the resistor. Substitute Ohm’s law in it. Thus, you get a maximum current that can flow through a resistor A. Divide this value of current by 2 and get maximum current that can flow through three resistors.
Formula used:
$P\quad =\dfrac { { V }^{ 2 } }{ R }$
$V\quad =\quad IR$
Complete step by step answer:
Given: R= $2\Omega$,
Power (P) = 18W
Power dissipated by the resistor is given by,
$P\quad =\dfrac { { V }^{ 2 } }{ R }$ …(1)
where, P: Power dissipated by the resistor
V: Voltage across the resistor
R: Value of Resistance
According to Ohm’s Law,
$V\quad =\quad IR$ …(2)
By substituting the equation. (2) in equation. (1) we get,
$P\quad =\quad { I }^{ 2 }R$
Rearranging the above equation we get,
${ I }^{ 2 }\quad =\quad \dfrac { P }{ R }$
$\therefore \quad I\quad =\quad \sqrt { \dfrac { P }{ R } }$ …(3)
Now, by substituting values in equation.(3) we get,
$\therefore \quad I\quad =\quad \sqrt { \dfrac { 18 }{ 2 } }$
$\therefore \quad I\quad =\quad 3A$
Thus, maximum current passing through resistor A is 3A.
The resistors B and C are connected in parallel hence, the current coming from resistor A is equally divided between resistors B and C.
Therefore, Current through B= Current through C = $\dfrac { 1 }{ 2 }$ Current through A
$ \therefore$ Current through B= Current through C = $ \dfrac { 1 }{ 2 } \quad \times \quad 3$
$ \therefore$ Current through B= Current through C = 1.5A
Hence, the maximum current that can flow through three resistors is 1.5A.
Note:
As more and more resistors are connected in parallel in a circuit, the equivalent resistance of the circuit decreases while the total current of the circuit decreases. If the resistors B and C were connected in series then the current flowing through each resistor would be the same.
Formula used:
$P\quad =\dfrac { { V }^{ 2 } }{ R }$
$V\quad =\quad IR$
Complete step by step answer:
Given: R= $2\Omega$,
Power (P) = 18W
Power dissipated by the resistor is given by,
$P\quad =\dfrac { { V }^{ 2 } }{ R }$ …(1)
where, P: Power dissipated by the resistor
V: Voltage across the resistor
R: Value of Resistance
According to Ohm’s Law,
$V\quad =\quad IR$ …(2)
By substituting the equation. (2) in equation. (1) we get,
$P\quad =\quad { I }^{ 2 }R$
Rearranging the above equation we get,
${ I }^{ 2 }\quad =\quad \dfrac { P }{ R }$
$\therefore \quad I\quad =\quad \sqrt { \dfrac { P }{ R } }$ …(3)
Now, by substituting values in equation.(3) we get,
$\therefore \quad I\quad =\quad \sqrt { \dfrac { 18 }{ 2 } }$
$\therefore \quad I\quad =\quad 3A$
Thus, maximum current passing through resistor A is 3A.
The resistors B and C are connected in parallel hence, the current coming from resistor A is equally divided between resistors B and C.
Therefore, Current through B= Current through C = $\dfrac { 1 }{ 2 }$ Current through A
$ \therefore$ Current through B= Current through C = $ \dfrac { 1 }{ 2 } \quad \times \quad 3$
$ \therefore$ Current through B= Current through C = 1.5A
Hence, the maximum current that can flow through three resistors is 1.5A.
Note:
As more and more resistors are connected in parallel in a circuit, the equivalent resistance of the circuit decreases while the total current of the circuit decreases. If the resistors B and C were connected in series then the current flowing through each resistor would be the same.
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