Circuit Current

Electricity and Magnetism Level 1

The picture above depicts a 17 V 17\text V battery arranged into a circuit of five resistors ( R R ). In the circuit, R I = 5 Ω R_I = 5\Omega , R I I = 7 Ω R_{II} = 7\Omega , R I I I = 4 Ω R_{III} = 4\Omega , R I V = 2 Ω R_{IV} = 2\Omega , and R V = 3 Ω R_V = 3\Omega . Determine the current ( I I ) of the circuit.

Give your answer to two decimal places.

David's Electricity Set


The answer is 1.21.

View wiki
David Hontz by David Hontz , 26, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

David Hontz
May 23, 2016

Relevant wiki: Series and parallel resistors

The key is to simplify the resistors by noticing which are in series and which are in parallel. R s e r i e s 1 = R I I I + R I V = 4 + 2 = 6 Ω 1 R p a r a l l e l 1 = 1 R V + 1 R s e r i e s 1 = 1 3 + 1 6 = 1 2 ⇒ R p a r a l l e l 1 = 2 Ω R c i r c u i t = R I + R I I + R p a r a l l e l 1 = 5 + 7 + 2 = 14 Ω R_{series \space 1} = R_{III} + R_{IV} = 4+2=\boxed{6Ω} \\ \frac{1}{R_{parallel \space 1}} =\frac{1}{R_{V}} + \frac{1}{R_{series \space 1}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2} \Rightarrow R_{parallel \space 1} = \boxed{2Ω} \\ R_{circuit} = R_I + R_{II} + R_{parallel \space 1} = 5+7+2=\boxed{14Ω} \\

V = I R I c i r c u i t = V c i r c u i t R c i r c u i t = 17 V 14 Ω ≈ 1.21 A V = IR \\ I_{circuit} = \frac{V_{circuit}}{R_{circuit}} = \frac{17V}{14Ω} \approx \boxed{1.21A}

16 Helpful 0 Interesting 0 Brilliant 0 Confused

why is the answer not negative? -1.21A?

Peter Lau - 4 years, 11 months ago

Log in to reply

I'm not sure where the negative would come from. If you could explain the math you did to get a negative, I would perhaps be better able to answer your question. Otherwise, because the overall circuit's resistance and voltage values are positive; the current must also be positive as well.

David Hontz - 4 years, 11 months ago

Log in to reply

It's actually due to the direction of current as indicated in the diagram, the actual direction of current is opposite to it. Hence, it should be negative.

Akshat Sharda - 3 years, 9 months ago

Log in to reply

@Akshat Sharda – Ah thanks. Those who answered -1.21 have been marked correct. I have updated the image and reversed the polarity of the battery.

Calvin Lin Staff - 3 years, 9 months ago

Hi David, as our decimal evaluation allows for a 2% margin of error, I've updated the answer to 3 significant figures / 2 decimal places.

This problem really appealed​ to the E+M folks!

Calvin Lin Staff - 5 years ago

Log in to reply

I see. Thanks for letting me know. I'm glad the question was well liked!

David Hontz - 5 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...