how would you connect three resistors of 12 ohm each so that the combination has a resistance of
a)18 ohm
b)8 ohm
Answers
Explanation:
a) 12 ohm
For 12 ohm we will connect firstly two resistors in parallel and the combination of these resistors in series with the third resistor
Firstly two resistors in parallel
\implies\sf{\dfrac{1}{\ R_{p}}=\dfrac{1}{8}+\dfrac{1}{8}}⟹
R
p
1
=
8
1
+
8
1
\implies\sf{\ R_{p}=\dfrac{8\times 8}{8+8}}⟹ R
p
=
8+8
8×8
\implies\sf{\ R_{p}=\dfrac{64}{16}}⟹ R
p
=
16
64
\implies\sf{\ R_{p}=\dfrac{\cancel{64}}{\cancel{16}}}⟹ R
p
=
16
64
\implies\sf{\ R_{p}=4\:ohm}⟹ R
p
=4ohm
Now in series with third resistor
\implies\sf{\ R_{s}=4+8}⟹ R
s
=4+8
\implies\sf{\ R_{s}=12\:ohm}⟹ R
s
=12ohm
So , there is an equivalent resistance of 12 ohm
b) 5.3 ohm
For 5.3 ohm we will connect firstly two resistors in series and the combination of these resistors in parallel with the third resistor
Firstly two resistors in series
\implies\sf{\ R_{s}=8+8}⟹ R
s
=8+8
\implies\sf{\ R_{s}=16\:ohm}⟹ R
s
=16ohm
Now connect this resultant with the III resistor in parallel
\implies\sf{\dfrac{1}{\ R_{p}}=\dfrac{1}{16}+\dfrac{1}{8}}⟹
R
p
1
=
16
1
+
8
1
\implies\sf{\ R_{p}=\dfrac{16\times 8}{16+8}}⟹ R
p
=
16+8
16×8
\implies\sf{\ R_{p}=\dfrac{128}{24}}⟹ R
p
=
24
128
\implies\sf{\ R_{p}=5.3\:ohm}⟹ R
p
=5.3ohm
So , there is an equivalent
5.3ohm