Question.
What is
(a) the highest,
(b) the lowest total resistance that can be secured by combinations of four coils of resistances $4 \Omega, 8 \Omega, 12 \Omega, 24 \Omega$ ?
What is
(a) the highest,
(b) the lowest total resistance that can be secured by combinations of four coils of resistances $4 \Omega, 8 \Omega, 12 \Omega, 24 \Omega$ ?
solution:
(a) The highest resistance is secured when all the resistors are connected in series. The equivalent resistance is given by
$R_{g}=4 \Omega+8 \Omega+12 \Omega+24 \Omega=48 \Omega$
(2) The lowest resistance is secured when all the four coils are connected in parallel. The equivalent resistance is given by,
$\frac{1}{R_{e}}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}=\frac{6+3+2+1}{24}=\frac{12}{24}=\frac{1}{2} \quad$ or $\quad R_{e}=2 \Omega$
(a) The highest resistance is secured when all the resistors are connected in series. The equivalent resistance is given by
$R_{g}=4 \Omega+8 \Omega+12 \Omega+24 \Omega=48 \Omega$
(2) The lowest resistance is secured when all the four coils are connected in parallel. The equivalent resistance is given by,
$\frac{1}{R_{e}}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}=\frac{6+3+2+1}{24}=\frac{12}{24}=\frac{1}{2} \quad$ or $\quad R_{e}=2 \Omega$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.