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If the battery provides a 12 V electromotive force and the resistors are each 10 ohm , what is the current going through the resistor in region A in Amps?enter image description here

I assume I need Ohm's law: V = IR.

I would think for branch A, I would plug in the values 12 = I * 10

because there is only one resistor in branch A. This gives me I = 1.2, which is incorrect.

Where am I going wrong?

tyger2020
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    Why do you assume that the potential difference between two ends of A is 12? – Sangchul Lee Feb 24 '18 at 02:23
  • Lookup series and parallel resistance e.g. here. – dxiv Feb 24 '18 at 02:25
  • The current has to flow through all three regions. You can't ignore the effects regions B and C will have on the current. Use parallel and series equivalent resistance methods to get a hold of the current and voltage values you need. – NovaDenizen Feb 24 '18 at 02:28

3 Answers3

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12V is dropped across the whole resistor network, not resistor $R_a$ (the resistor in region A). Less voltage is dropped across $R_a$.

Using Kirchoff's voltage law you can write:

$$V_a + V_b + V_c = 10$$

Using Kirchoff's current law you can write:

$$I_a = I_{batt}$$ $$I_a - I_{b1} - I_{b2} = 0$$ $$I_{b1}+I_{b2}-I_{c1}-I_{c2}-I_{c3}=0$$ $$I_{c1}+I_{c2}+I_{c3}=I_{batt}$$

And Ohm's law lets you write:

$$V_a=I_aR_a$$ $$V_{b}=I_{b1}R_{b1}$$ etc.

From that system of linear equations, and knowing that all the $R$'s are 10 ohms, you can solve for all the $V$'s and $I$'s.

Andy Walls
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Overall resistance in the circuit = $10 + \frac{10}2 + \frac{10}3 = \frac{55}3$ ohm.

Current $I = \frac VR = \frac{12}{\frac{55}3} = \frac{36}{55}$ ampere

Deepak
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Note that total resistance is equivalent to $R + (R^{-1} + R^{-1})^{-1} + (R^{-1} + R^{-1} + R^{-1})^{-1}$

Which means that $I_{a} = \frac{V}{R + (R^{-1} + R^{-1})^{-1} + (R^{-1} + R^{-1} + R^{-1})^{-1}}$

bigfocalchord
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