Question

(a) What is the resistance of a \(1.00 \times {10^2}\;\Omega \), a \(2.50\;{\rm{k}}\Omega ,\)and a \({\bf{4}}{\bf{.00}}\;{\bf{k\Omega }}\) resistor connected in series? (b) In parallel?

Short answer

(a) The resistance when connected in series give \(6.6\;{\rm{k}}\Omega \).

(b) The resistance when connected in parallel give \(93.9\;\Omega \).

Step-by-step solution

Given Data

The values of the three resistances given, are-

\(\begin{align}{}{R_1} = 100\;\Omega \\{R_2} = 2500\;\Omega \\{R_3} = 4000\;\Omega \end{align}\)

Definition of resistance in series and parallel

Resistance in series: equivalent resistance in series combination can be obtained by adding the resistance of all the resistors.

Resistance in parallel: in parallel combination, the reciprocal of equivalent resistance can be calculated by adding the reciprocal of all the resistances.

Finding resistance when connected in series 

(a)

When three resistors are connected in series,

\({R_{eq}} = {R_1} + {R_2} + {R_3}\)

\(\begin{align}{}{R_{eq.}} &= 100\;\Omega + 2500\;\Omega + 4000\;\Omega \\ &= 6600\;\Omega \\ &= 6.6\;{\rm{k}}\Omega \end{align}\)

The equivalent resistance in series combination is given as- \(6.6\;{\rm{k\Omega }}\)

Finding resistance when connected in parallel

(b

when the resistors are connected in parallel,

\(\begin{align}{}\frac{1}{{{R_{eq}}}} &= \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}}\\ &= \frac{1}{{100\;\Omega \;}} + \frac{1}{{2500\;\Omega }} + \frac{1}{{4000\;\Omega }}\\ &= \left( {\frac{{200 + 8 + 5}}{{20000}}} \right)\;{\Omega ^{ - 1}}\\ &= \frac{{213}}{{20000}}\;\;{\Omega ^{ - 1}}\end{align}\)

Further,

\(\begin{align}{}{R_{eq}} &= \frac{{20000}}{{213}}\;\Omega \\ &= 93.9\;\Omega \end{align}\)

The equivalent resistance in parallel combination is given as- \(93.9\;\Omega \).