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Class -10th ​

Answer 1

Solution :

(a) Total resistance of circuit :

• Suppose we replace the parallel resistors R_1 and R_2 by an equivalent resistor of resistance R'. Similarly we replace the parallel resistors R_3 ,R_4 and R_5 by an equivalent single resistor of Resistance R". Then using :

$\sf {\dfrac {1}{R_p} = \dfrac {1}{R_1} + \dfrac {1}{R_2} + \dfrac {1}{R_3} }$

ㅤㅤㅤㅤ$\small {\mathfrak {\underline {\red { Substituting \: the \: values \: in \: formula \: :- }}}}$

$| \implies \sf {\dfrac {1}{R'} = \dfrac {1}{10} + \dfrac {1}{40} }$

$| \implies \sf { \dfrac {1}{R'} = \dfrac { 5}{40} }$

$| \implies \sf { R'= 8 \: ohm}$

$| \implies \sf { \dfrac {1}{R"} = \dfrac {1}{30} + \dfrac {1}{20} + \dfrac {1}{60} = \dfrac {6}{60} }$

$| \implies \sf { R"=10 \: ohm}$

Total resistance = R' + R"

ㅤㅤㅤㅤㅤㅤㅤㅤ= 8 + 10

ㅤㅤㅤㅤㅤㅤㅤㅤ= 18 ohm

(b) Total Current flowing through circuit :

$\sf {Ohm's \: Law : \: I = \dfrac {V}{R} }$

$\implies \sf { I = \dfrac {12 \: V }{18 \: ohm}}$

$\implies \sf { I = 0.67 \: A}$

Answer 2

Solution :

(a) Total resistance of circuit :

• Suppose we replace the parallel resistors R_1 and R_2 by an equivalent resistor of resistance R'. Similarly we replace the parallel resistors R_3 ,R_4 and R_5 by an equivalent single resistor of Resistance R". Then using :

$\sf {\dfrac {1}{R_p} = \dfrac {1}{R_1} + \dfrac {1}{R_2} + \dfrac {1}{R_3} }$

ㅤㅤㅤㅤ$\small {\mathfrak {\underline {\pink{ Substituting \: the \: values \: in \: formula \: :- }}}}$

$| \implies \sf {\dfrac {1}{R'} = \dfrac {1}{10} + \dfrac {1}{40} }$

$| \implies \sf { \dfrac {1}{R'} = \dfrac { 5}{40} }$

$| \implies \sf { R'= 8 \: ohm}$

$| \implies \sf { \dfrac {1}{R"} = \dfrac {1}{30} + \dfrac {1}{20} + \dfrac {1}{60} = \dfrac {6}{60} }$

$| \implies \sf { R"=10 \: ohm}$

Total resistance = R' + R"

ㅤㅤㅤㅤㅤㅤㅤㅤ= 8 + 10

ㅤㅤㅤㅤㅤㅤㅤㅤ= 18 ohm

(b) Total Current flowing through circuit :

$\sf {Ohm's \: Law : \: I = \dfrac {V}{R} }$

$\implies \sf { I = \dfrac {12 \: V }{18 \: ohm}}$

$\implies \sf { I = 0.67 \: A}$