Question

Five resistors are connected in a circuit as shown. Find the ammeter reading when the circuit is closed.

Answer 1

Answer: 1 Ampere

Given data states the 5 resistors connected with the value,

$R_1$ = 3 Ω; $R_2$ = 3 Ω ; $R_3$ = 3 Ω; $R_4$ = 0.5 Ω and $R_5$ = 0.5 Ω.

As shown,  the $R_1$ and $R_2$ are in series connection, therefore,

R' =  $R_1 + R_2$ = 3 + 3 = 6 Ω

And then R' and $R_3$ are now in parallel connection, thereby

$\frac { 1 }{ R'' } =\frac { 1 }{ R' } +\frac { 1 }{ R_{ 3 } }$

$\Rightarrow \frac { 1 }{ R'' } =\frac { 1 }{ 6 } +\frac { 1 }{ 3 }$

$\Rightarrow R''=2$Ω

And then R'' with $R_4$ and $R_5$ are in series, thereby

Total $R = R'' + R_4 + R_5$

$\Rightarrow Total\quad R = 0.5 + 2 + 0.5$

$\Rightarrow Total\quad R = 3$ Ω.

Therefore,

Current $I=\frac { V }{ R }$

$\Rightarrow I=\frac { 3 }{ 3 }$ =1 Ampere.

Answer 2

Answer:

Explanation:

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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