Question

Find the (i) Maximum and (ii) Minimum Equivalent resistance, when 400Ω, 800 Ω, 1000 Ω and 2000 Ω resistors are connected?

Answer 1

Answer:

$\boxed{ \bold{ \pink{Maximum \: resistance = 4200 ohm}}} \\ \boxed{\bold{ \blue{Minimum \: resistance = 190.47ohm}}}$

Explanation:

$\bold{ \underline{ \red{Given}}} \longrightarrow \\ four \: resistances \: 400ohm, \: 800hm ,\: 1000ohm ,\: 2000ohm \\ are \: connected \: to \: each \: other$

$\bold{ \underline{ \red{To \: find}}} \longrightarrow \\ maximum \: and \: minimum \: value \: of \: resultant \: resistances$

$\bold{ \underline{ \red{Concept \: used}}} \longrightarrow \\ we \: know \: that \: resultant \: resistance \: is \: \\ maximum \: when \: we \: connect \: resistances \: in \: series \\ and \: minimum \: when \: we \: connect \\ \: resistances \: in \: parallel$

$\bold{ \underline{ \red{solution}}} \longrightarrow \\ we \: get \: maximum \: combined \: resistance \\ \: when \: we \: connect \: resistances \: in \: series \\ maximum \: resultant \: resistane \\ = 400 + 800 + 1000 + 2000 \\ = 4200ohm \\ now \\ \:for \: minimum \: resultant \: resistance \\ \: we \: connect \: all \: resistances \: in \: parallel \: \\ \dfrac{1}{R} = \dfrac{1}{400} + \dfrac{1}{800} + \dfrac{1}{1000} + \dfrac{1}{2000} \\ = \dfrac{10 + 5 + 4 + 2}{4000} \\ = \dfrac{21}{4000} \\ R = \dfrac{4000}{21} \\ R = 190.47ohm \\ minimum \: resultant \: resistances = 190.47ohm$