Physics, asked by gouravsaini7119, 1 year ago

Five resistance, each of 5 Ω, are connected as shown in figure. Find the equivalent resistance between points (1) A and B, (2) A and C.(a) (1) 7.5 Ω (2)2.25 Ω(b) (1) 5 Ω (2) 2.50 Ω(c) (1) 2.5 Ω (2) 3.13 Ω(d) (1) 3 Ω (2) 2.50 Ω

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Answered by amy001
26

Hey here's your answer..hope it helps ..plz mark it as brainliest

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Answered by nain31
42
 \huge \bold{RESISTANCE}

Resistance is the obstruction offered in the path of the current.

◼When resistors are connected in series the total resistance is equal to

 \boxed{R_s = R_1 + R_2 + n \: resistors }

◼When resistors are connected in parallel the total resistance is equal to

 \boxed{\dfrac{1}{R_s} = \dfrac{1}{R_1}+\dfrac{1}{R_2} + n resistors }

 \huge \bold{IN \: THE \: FIGURE}

 \bold{RESISTANCE \: BETWEEN \: A \: AND \: B \: WILL \: BE}

Let the point from which the wire in divided into three parallel parts be \bold{ P} and the point where the unite be  \bold{ Q} and the resistor in middle be  \bold{R_p}

So net resistance between P and Q is

 \mathsf{R_{s1} = 5 + 5 }

 \boxed{R_{s1} = 10 \: ohm }

So net resistance between P ,C and Q is

 \mathsf{R_{s2} = 5 + 5 }

 \boxed{R_{s2} = 10 \: ohm }

Since,  \bold{R_p , R_{s1} \: and R_{s2} \: are \: in \: parallel }

So, total resistance R will be

 \boxed{\dfrac{1}{R} = \dfrac{1}{R_{s1}}+ \dfrac{1}{R_p}+\dfrac{1}{R_{s2}}}

 \mathsf{\dfrac{1}{R} = \dfrac{1}{10}+ \dfrac{1}{5}+\dfrac{1}{10}}

 \mathsf{\dfrac{1}{R} = \dfrac{1+1+2}{10}}

 \mathsf{\dfrac{1}{R} = \dfrac{4}{10}}

 \mathsf{R = \dfrac{10}{4}}

 \huge \boxed{\mathsf{R = 2.5 \: ohm}}

 \bold{RESISTANCE \: BETWEEN \: A \: AND \: C \: WILL \: BE}

The resistor between A, B and C are in series,

So net resistance between P and Q is

 \mathsf{R_{s3} = 5 + 5 }

 \boxed{R_{s3} = 10 \: ohm }

\bold{Resistor R_{s3} \: and \: R_p \: are \: in \: parallel}

So, total resistance R_2 will be

 \boxed{\dfrac{1}{R_2} = \dfrac{1}{R_{s3}}+ \dfrac{1}{R_p}}

 \mathsf{\dfrac{1}{R_2} = \dfrac{1}{10}+ \dfrac{1}{5}}

 \mathsf{\dfrac{1}{R_2} = \dfrac{1 + 2}{10}}

 \mathsf{\dfrac{1}{R_2} = \dfrac{3}{10}}

 \mathsf{R_2 = \dfrac{10}{3}}

 \huge \boxed{\mathsf{R = 3.33 \: ohm}}

nain31: no useless comments :)
nain31: :)
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