circuit q pls help yr 11 (1 Viewer)

Vaibhav123456

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I dont remember circuits too well but i believe you minus the voltage drop from the first resistor. So V = 15 - 4 = 11V. Then you use this voltage for the second resistor and the current from part a.
 

yashbb

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I dont remember circuits too well but i believe you minus the voltage drop from the first resistor. So V = 15 - 4 = 11V. Then you use this voltage for the second resistor and the current from part a.
thank you so so much you are correct.
 

notme123

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Don't worry too much about circuits if you're stuck because it doesn't come back in detail in year 12. I personally sucked at this and was happy to not see it return.
 
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yashbb

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Don't worry too much about circuits if you're stuck because it doesn't come back in detail in year 12. I personally sucked at this as was happy to not see it return.
ok ok ok tysm, does magnetism come back?
 

dasfas

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Yeah what Vaibhav said :)

Voltage across the 5Ω resistor is 4V, hence the current i = 4/5A. Since the resistors are in series, you know the current through R is 4/5A. We can use KVL to determine that the voltage is 11V across R, hence re-arranging R = V/i, we get an expression for the unknown resistor R.

Screen Shot 2021-09-02 at 9.46.41 pm.png
 

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