Question

and 22 ω. determine the total resistance.

1. find the total resistance of each of the following groups of series resistors:

(a) 560 ω and 1000 ω
(b) 47 ω and 56 ω
(c) 1.5 kω, 2.2 kω, and 10 kω
(d) 1.0 mω, 470 kω, 1.0 kω, 2.2 mω

Explanation:

Step1: Recall series - resistance formula

For resistors in series, the total resistance $R_{total}$ is given by $R_{total}=R_1 + R_2+\cdots+R_n$.

Step2: Calculate (a)

Given $R_1 = 560\Omega$ and $R_2=1000\Omega$. Then $R_{total}=R_1 + R_2=560\Omega+ 1000\Omega = 1560\Omega$.

Step3: Calculate (b)

Given $R_1 = 47\Omega$ and $R_2 = 56\Omega$. Then $R_{total}=R_1+R_2=47\Omega + 56\Omega=103\Omega$.

Step4: Calculate (c)

First convert to ohms: $1.5k\Omega=1.5\times10^{3}\Omega$, $2.2k\Omega = 2.2\times10^{3}\Omega$, $10k\Omega=10\times10^{3}\Omega$. Then $R_{total}=1.5\times10^{3}\Omega+2.2\times10^{3}\Omega + 10\times10^{3}\Omega=(1.5 + 2.2+10)\times10^{3}\Omega=13.7\times10^{3}\Omega = 13.7k\Omega$.

Step5: Calculate (d)

First convert to ohms: $1.0M\Omega=1.0\times10^{6}\Omega$, $470k\Omega=470\times10^{3}\Omega = 0.47\times10^{6}\Omega$, $1.0k\Omega=1.0\times10^{3}\Omega=0.001\times10^{6}\Omega$, $2.2M\Omega=2.2\times10^{6}\Omega$. Then $R_{total}=(1.0 + 0.47+0.001 + 2.2)\times10^{6}\Omega=3.671\times10^{6}\Omega=3.671M\Omega$.

Answer:

(a) $1560\Omega$
(b) $103\Omega$
(c) $13.7k\Omega$
(d) $3.671M\Omega$