In the figure shown, what is the current (in Ampere) drawn from the battery? You are given:
$R_{1} = 15 Ω, R_{2} = 10 Ω, R_{3} = 20 Ω, R_{4} = 5 Ω, R_{5} = 25 Ω, R_{6} = 30 Ω, E = 15 V$

Question

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given: R 1 = 15 &#160; &#937; &#160; , &#160; R 2 = 10 &#160; &#937; &#160; , &#160; R 3 = 20 &#160; &#937; &#160; , &#160; R 4 = 5 &#160; &#937; , &#160; R 5 = 25 &#160; &#937; , &#160; R 6 = 30 &#160; &#937; , &#160; E = 15 &#160; V

Aicharya · Accepted Answer

The equivalent resistance R e q of this given network is, R e q = 15 + 25 3 + 30 = 45 + 25 + 90 3 = 160 3 By applying ohm's law, the Current i through the battery is i = 15 160 3 = 15 &#215; 3 160 &#8756; i = 9 32 &#160; A

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given: R 1 = 15  …

Solution

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given: R 1 = 15 &#160;…

Solution

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given: R 1 = 15 …