What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answers
Answer:
a.) 48
b.) 2
Explanation:
a.) Resistance in series is greater than the greatest so , highest resistance :
R = 4+8+12+24 = 48 ohm
b.) Resistance in parallel is smaller than the smallest. so, lowest resistance :
1/ R = 1/4+1/8+1/12+1/24 = 6+3+2+1/24 = 12/24 ohm
R = 2 ohm
Explanation:
Given:−
\rm\:R_1=4\text{\O}megaR
1
=4Ømega
\rm\:R_2=8\text{\O}megaR
2
=8Ømega
\rm\:R_3=12\text{\O}megaR
3
=12Ømega
\rm\:R_4=24\text{\O}megaR
4
=24Ømega
{\bold{\orange{\underline{\pink{To}\:\red{Fi}\purple{nd}\green{:-}}}}}
ToFind:−
\rm\:The\:highest\:resistance=?Thehighestresistance=?
\rm\:The\:lowest\:resistance=?Thelowestresistance=?
{\bold{\blue{\underline{\red{So}\purple{lut}\green{ion}\orange{:-}}}}}
Solution:−
\rm\:(a)(a) The highest resistance can be secured by connecting,all four coil in series.
\tt\:In\:this\:case:-Inthiscase:−
\rm\:R=R_1+R_2+R_3+R_4R=R
1
+R
2
+R
3
+R
4
\rm\:R=4+8+12+24R=4+8+12+24
\rm\:R=48\text{\O}megaR=48Ømega
Thus,the highest resistance which can be secured is 48 Ω.
\rm\:(b)(b) The lowest resistance can be secured by connecting all four coils in parallel.
\tt\:In\:this\:case:-Inthiscase:−
\rm\:\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}
R
1
=
R
1
1
+
R
2
1
+
R
3
1
+
R
4
1
\rm\:\dfrac{1}{R}=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}
R
1
=
4
1
+
8
1
+
12
1
+
24
1
\rm\:\dfrac{1}{R}=\dfrac{6+3+2+1}{24}
R
1
=
24
6+3+2+1
\rm\:\dfrac{1}{R}=\cancel\dfrac{12}{24}
R
1
=
24
12
\rm\:\dfrac{1}{R}=\dfrac{1}{2}
R
1
=
2
1
\rm\:R=2\text{\O}megaR=2Ømega
Thus,the lowest resistance which can be secured is 2 Ω.