Oxford PAT 2009, Question 25

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6 thoughts on “Oxford PAT 2009, Question 25

  1. I’m sorry , I guess I wasn’t clear in my question. I’m meant to ask why do you divide 5 by 5000 to obtain current I.e, why the resistance is taken to be 5000?
    Also, I didn’t get the calculations you did to work out the voltage between A and B, so can you please explain them.
    Thanks

    1. Hi Arjun

      OK that’s great — I’ll go through the details line by line.

      1) the first line that relates I3 and I4. I3 flows through a single resistor of resistance R, and I4 flows through two resistors in series, which have a resistance 2R (because that’s what you get when you connect two Rs in series). Because the single resistor and the two series-connected resistors are connected in parallel, the voltage across them must be the same, so because V=IR, we can see that I3 must be equal to 2 I4.

      2) the second line that relates I2, I3 and I4. Because the sum of the currents flowing into a point must be the same as the sum of the currents flowing out of it, we can see that I2 must be equal to I3 + I4. So by using the previous equation I3 = 2 I4, we can see that I2 = 2 I4 + I4 so I2 = 3 I4.

      3) the third line referring to Vab. The current I2 is flowing through a 1K resistor, and the current I3 is flowing through a 1K resistor, and the voltage across these two resistors adds up to Vab. This is where we get the expression Vab = 1000(I2 + I3), and from (1) and (2) we can see that this means Vab = 5000 I4. But we can also see that the current I1 flows through a 1K resistor, and the voltage across that resistor is Vab, so Vab = 1000 I1. And therefore I1 = 5 I4.

      4) the fourth line. The total current flowing is I1 + I2, so the effective resistance is Vab / (I1 + I2). We know that I2 = 3 I4, and I1 = 5 I4, therefore I1 = 8 I4 / 5. So the resistance is 5 Vab / 8 I1. But we can also that I1 flows through a 1K resistor and the voltage across that resistor is Vab, so Vab / I1 = 1000. Therefore the effective resistance is 5 x 1000 / 8, which is 625 ohms.

      5) the fifth line, deriving Vab. The current in the whole circuit flows through the resistor network, with resistance 625 and through the battery, with resistance 125. So the voltage across the resistor network is the battery voltage (which is 6V) multiplied by 625 / (125 + 625), which is 5V.

      6) the sixth line, deriving I4. We can see that I1 = 5 / 1000, and we know that I1 = 5 I4, therefore I4 = 5 / 5000. So I4 = 1mA.

        1. The resistance of the resistor network is 625Ω and the resistance of the battery is 125Ω, so the total resistance is (625+125)Ω. This means that the current flowing through the battery-resistor network circuit is 6/(625+125)A. Therefore the voltage across the resistor network is 625 x 6/(625+125), which equals 5V.

  2. Hi Arjun, thanks for your question.

    The current is the current between C and D, which is I4. In the working above, we have worked out that I4 is 5/5000 amps, which is 1mA.

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