Question

7) 20 resistors of 4 ohm each are connected in parallel. Calculate the effective
resistance?(2)

Answer 1

Answer:

0.2 Ω

Explanation:

As per the provided information in the given question, we have :

• 20 resistors of 4Ω each are connected in parallel.

We've been asked to calculate the effective resistance.

As we know that, when the resistors are connected in parallel combination, effective resistance is given by ;

$\longrightarrow\underline{ \boxed{\bf{\dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots \dfrac{1}{R_n} }}}$

$\mapsto \bf{\dfrac{1}{R_P} } \rm{= \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dfrac{1}{R_4} + \dfrac{1}{R_5} + \dfrac{1}{R_6}} \\ \rm{ + \dfrac{1}{R_7} + \dfrac{1}{R_8} + \dfrac{1}{R_9} + \dfrac{1}{R_{10}} + \dfrac{1}{R_{10}} + } \\ \rm{\dfrac{1}{R_{11}} + \dfrac{1}{R_{12}} + \dfrac{1}{R_{13}} + \dfrac{1}{R_{14}} + \dfrac{1}{R_{15}} + \dfrac{1}{R_{16}}} \\ \rm{\dfrac{1}{R_{17}} +\dfrac{1}{R_{18}} + \dfrac{1}{R_{19}} + \dfrac{1}{R_{20}} }$

Now, all the resistors have same value that is 4Ω. So, we can write it as,

$\mapsto \bf{\dfrac{1}{R_P} } \rm{= \bigg(\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4}} \\ \rm{ + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + } \\ \rm{\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4}} \\ \rm{\dfrac{1}{4} +\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} } \bigg) \: \Omega$

(Terms have broken the terms into different lines for readability.)

$\mapsto \rm{\dfrac{1}{R_P} } \rm{=\bigg( \dfrac{1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1}{4} } \bigg ) \: \Omega$

Performing addition in the numerator.

$\mapsto \rm{\dfrac{1}{R_P} } \rm{=\bigg( \dfrac{20}{4} } \bigg ) \: \Omega$

$\mapsto \rm{R_P} \rm{ = \dfrac{4}{20} } \: \Omega$

$\mapsto \rm{R_P} \rm{ = \dfrac{1}{5} } \: \Omega$

$\mapsto \underline{ \boxed{ \bf{{R_P}= 0.2 } \: \Omega}}$

Therefore, effective resistance is 0.2 Ω.

Answer 2

Answer:

0.2 Ohms

Explanation:

1/R(eff) = 1/R1 + 1/R2 +... + 1/R20

1/R(eff) = 20(1÷4)

R(eff) = 4/20 = 1/5 = 0.2 Ohms