Question

five resistors are connected in a circuit as shown in figure find the ammeter reading when circuit is closed.
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Answer 1

Answer:

Given data states the 5 resistors connected with the value,

= 3 Ω; = 3 Ω ; = 3 Ω; = 0.5 Ω and = 0.5 Ω.

As shown,  the and are in series connection, therefore,

R' =   = 3 + 3 = 6 Ω

And then R' and are now in parallel connection, thereby

Ω

And then R'' with and are in series, thereby

Total

Ω.

Therefore,

Current

=1 Ampere.

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Answer 2

$\boxed{\underline{\underline{\mathcal{ANSWER:-}}}}$

$\rm \: R_{1} \: and \: R_{2} \: are \: in \: series,$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \: R_{S1} \: = \: R_{1} \: + R_{2}=3+3=6 \Omega \:$

$\rm \: R_{S1} \: and \: R_{3} \: are \: in \: parallel.$

$\therefore \: \: \: \: \: \: \: \: \: \: \rm \: \frac{1}{R_P} = \frac{1}{R_{S1}} \: + \frac{1}{R_3}$

$\rm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: \frac{1}{6} + \frac{1}{3} = \frac{1}{2}$

$\implies \: \: \: \: \: \rm \: R{_P} = 2 \: \Omega$

$\rm \: R{_4}, \: R{_P} \: and \: R{_5} \: are \: in \: series.$

$\therefore \: \: \: \: \: \: \rm \: R{_S} = R{_4} + R{_P} + R{ _ 5} = 0.5 + 2 + 0.5 = 3 \: \Omega$

$\rm \: Then, \: current, \: I = \frac{V}{R{_S}} = \frac{3}{3} = \boxed{ \underline{\blue{ \bf \: 1 A}}}$

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