find the total resistance if 1 ohm 2 ohm and 4 ohm connected in parallel
Answers
Answer:
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Explanation:
In parallel combination , the equivalent resistance of three resistors is given by :
\tt{\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}+ \dfrac{1}{R_3}}
R
p
1
=
R
1
1
+
R
2
1
+
R
3
1
Or The reciprocal of equivalent resistance in parallel is the sum of the reciprocal of individual resistances .
Given :
The three resistors are 2 Ω , 4 Ω and 6 Ω .
\tt{\dfrac{1}{R_p} = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} }
R
p
1
=
2
1
+
4
1
+
6
1
Taking out the LCM of 2 , 4 and 6
= \tt \dfrac{1 \times 6 + 1 \times 3 + 1 \times 2}{12}=
12
1×6+1×3+1×2
\tt \: = \dfrac{6 + 3 + 2}{12}=
12
6+3+2
\tt = \dfrac{11}{12}=
12
11
\tt {R_p = \dfrac{12}{11}}R
p
=
11
12
Therefore , the equivalent resistance is 1.09 Ω.
⇢In parallel combination , equivalent resistance is smaller than the smallest resistor . Here the smallest resistor was of 2 Ω and the equivalent resistance is smaller than it .
⇢In series combination , the sum of all resistances gives the total resistance . So , the resistance is always larger than the largest resistance .
Answer:
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Explanation:
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