show how you would connected three resistors ,each of resistance 6 G, so that the combination has a resistance of (1 ) 9 resistor, (2) 4 resistor.
Answers
Answer:
1) first add two resistors in parallel then add the resulting resistance in series with third resistor
1/6+1/6=1/3
3+6=9
2) first add two resistors in series then add the resulting resistance in parallel with third resistor
6+6=12
1/12+1/6=1/4
4
Answer:
Show how you would connected three resistors ,each of resistance 6 Ω, so that the combination has a resistance of (1) 9 Ω, (2) 4 Ω.
Answer:
1) Connect the parallel combination of two 6 Ω resistors with one 6 Ω in a series combination.
2) Connect the series combination of two 6 Ω resistors with one 6 Ω in a parallel combination.
Explanation:
1) To get the equivalent resistance of 9 Ω :
Refer to the attachment (i).
\rm R_1R
1
and \rm R_2R
2
are connected in parallel combination, so their combined resistance will be :
\begin{gathered} \longmapsto \rm { \dfrac{1}{R_{(1,2)}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} } \\ \end{gathered}
⟼
R
(1,2)
1
=
R
1
1
+
R
2
1
\begin{gathered} \longmapsto \rm { \dfrac{1}{R_{(1,2)}} =\Bigg ( \dfrac{1}{6} + \dfrac{1}{6}\Bigg ) \; \text{\O}mega } \\ \end{gathered}
⟼
R
(1,2)
1
=(
6
1
+
6
1
)Ømega
\begin{gathered} \longmapsto \rm { \dfrac{1}{R_{(1,2)}} =\Bigg ( \dfrac{1 + 1}{6}\Bigg ) \; \text{\O}mega } \\ \end{gathered}
⟼
R
(1,2)
1
=(
6
1+1
)Ømega
\begin{gathered} \longmapsto \rm { \dfrac{1}{R_{(1,2)}} =\Bigg ( \dfrac{2}{6}\Bigg ) \; \text{\O}mega } \\ \end{gathered}
⟼
R
(1,2)
1
=(
6
2
)Ømega
\begin{gathered} \longmapsto \rm { R_{(1,2)} =\Bigg ( \dfrac{6}{2}\Bigg ) \; \text{\O}mega } \\ \end{gathered}
⟼R
(1,2)
=(
2
6
)Ømega
\begin{gathered} \longmapsto \rm { R_{(1,2)} = 3 \; \text{\O}mega } \\ \end{gathered}
⟼R
(1,2)
=3Ømega
Now, the combined resistance of \rm R_1R
1
and \rm R_2R
2
is connected in series combination with \rm R_3R
3
. So, equivalent resistance of the circuit will be :
\begin{gathered} \longmapsto \rm { R_{(1,2,3)} = R_{(1,2)} + R_3 } \\ \end{gathered}
⟼R
(1,2,3)
=R
(1,2)
+R
3
\begin{gathered} \longmapsto \rm { R_{(1,2,3)} = ( 3 +6) \; \text{\O}mega } \\ \end{gathered}
⟼R
(1,2,3)
=(3+6)Ømega
\begin{gathered} \longmapsto \bf { R_{(1,2,3)} = 9 \; \text{\O}mega } \\ \end{gathered}
⟼R
(1,2,3)
=9Ømega
∴ If we connect the parallel combination of two 6 Ω resistors with one 6 Ω in a series combination, we get the equivalent resistance of 9 Ω.
2) To get the equivalent resistance of 4 Ω :
Refer to the attachment (ii).
\rm R_1R
1
and \rm R_2R
2
are connected in series combination, so their combined resistance will be :
\begin{gathered} \longmapsto \rm { R_{(1,2)} = R_1 + R_2 } \\ \end{gathered}
⟼R
(1,2)
=R
1
+R
2
\begin{gathered} \longmapsto \rm { R_{(1,2)} =( 6+6) \; \text{\O}mega} \\ \end{gathered}
⟼R
(1,2)
=(6+6)Ømega
\begin{gathered} \longmapsto \rm { R_{(1,2)} = 12 \; \text{\O}mega} \\ \end{gathered}
⟼R
(1,2)
=12Ømega
Now, the combined resistance of \rm R_1R
1
and \rm R_2R
2
is connected in parallel combination with \rm R_3R
3
. So, equivalent resistance of the circuit will be :
\begin{gathered} \longmapsto \rm { \dfrac{1}{R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)} } + \dfrac{1}{R_3} } \\ \end{gathered}
⟼
R
(1,2,3)
1
=
R
(1,2)
1
+
R
3
1
\begin{gathered} \longmapsto \rm { \dfrac{1}{R_{(1,2,3)}} =\Bigg ( \dfrac{1}{12} + \dfrac{1}{6}\Bigg ) \; \text{\O}mega } \\ \end{gathered}
⟼
R
(1,2,3)
1
=(
12
1
+
6
1
)Ømega
\begin{gathered} \longmapsto \rm { \dfrac{1}{R_{(1,2,3)}} =\Bigg ( \dfrac{1+2}{12} \Bigg ) \; \text{\O}mega } \\ \end{gathered}
⟼
R
(1,2,3)
1
=(
12
1+2
)Ømega
\begin{gathered} \longmapsto \rm { \dfrac{1}{R_{(1,2,3)}} =\Bigg ( \dfrac{3}{12} \Bigg ) \; \text{\O}mega } \\ \end{gathered}
⟼
R
(1,2,3)
1
=(
12
3
)Ømega
\begin{gathered} \longmapsto \rm { R_{(1,2,3)} =\Bigg ( \dfrac{12}{3}\Bigg ) \; \text{\O}mega } \\ \end{gathered}
⟼R
(1,2,3)
=(
3
12
)Ømega
\begin{gathered} \longmapsto \bf { R_{(1,2,3)} = 4 \; \text{\O}mega } \\ \end{gathered}
⟼R
(1,2,3)
=4Ømega
∴ If we connect the series combination of two 6 Ω resistors with one 6 Ω in a parallel combination, we get the equivalent resistance of 4 Ω.