Answer
64.8k+ views
Hint: From the diagram, it is clear that the two resistors are in parallel with the one resistor in the series. First calculate the equivalent resistance of the two parallel resistors using formula. Then with the equivalent resistance calculated, again find the total resistance with the series resistors using the formula.
Useful formula:
(1) The equivalent resistance of the resistors connected in parallel is given by
$r = \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$
Where $r$ is the equivalent resistor, ${R_1}$ is the first resistor and ${R_2}$ is the second resistor.
(2) The equivalent resistance of the resistors that connected in series,
${R_e} = r + R$
Where ${R_e}$ is the equivalent resistance.
Complete step by step solution:
It is given that the
Resistance of each resistor, ${R_1} = {R_2} = {R_3} = 9\,\Omega $
The equivalent resistance of the combination, $r = 13.5\,\Omega $
The formula of the equivalent resistance is taken.
$r = \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$
Substituting the values in the above step.
$r = \dfrac{{9 \times 9}}{{9 + 9}}$
By performing simple arithmetic operations.
$r = 4.5\,\Omega $
Using the formula (2) to calculate the equivalent resistance.
${R_e} = r + R$
Substituting the values in the above formula,
${R_e} = 4.5 + 9$
${R_e} = 13.5\,\Omega $
Hence the equivalent resistance of the given circuit is obtained as $13.5\,\Omega $.
Note: Normally adding the resistance in the circuit gives the equivalent resistance, if they are connected in series. Remember the formula of the equivalent resistance in parallel. The parallel circuit has the same voltage in each resistor and the series circuit has the same current in each circuit.
Useful formula:
(1) The equivalent resistance of the resistors connected in parallel is given by
$r = \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$
Where $r$ is the equivalent resistor, ${R_1}$ is the first resistor and ${R_2}$ is the second resistor.
(2) The equivalent resistance of the resistors that connected in series,
${R_e} = r + R$
Where ${R_e}$ is the equivalent resistance.
Complete step by step solution:
It is given that the
Resistance of each resistor, ${R_1} = {R_2} = {R_3} = 9\,\Omega $
The equivalent resistance of the combination, $r = 13.5\,\Omega $
The formula of the equivalent resistance is taken.
$r = \dfrac{{{R_1} \times {R_2}}}{{{R_1} + {R_2}}}$
Substituting the values in the above step.
$r = \dfrac{{9 \times 9}}{{9 + 9}}$
By performing simple arithmetic operations.
$r = 4.5\,\Omega $
Using the formula (2) to calculate the equivalent resistance.
${R_e} = r + R$
Substituting the values in the above formula,
${R_e} = 4.5 + 9$
${R_e} = 13.5\,\Omega $
Hence the equivalent resistance of the given circuit is obtained as $13.5\,\Omega $.
Note: Normally adding the resistance in the circuit gives the equivalent resistance, if they are connected in series. Remember the formula of the equivalent resistance in parallel. The parallel circuit has the same voltage in each resistor and the series circuit has the same current in each circuit.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)