Question 3 This question is worth 20 marks, and is designed...

Question 3 This question is worth 20 marks, and is designed to test your understanding of electrical circuits (Unit 10). (a) State in your own words Kirchhoff’s laws and Ohm’s law for electrical circuits. (Answers quoted from the module material or elsewhere will receive no marks.) (4 marks) (b) The circuit shown in Figure 1 can be used to measure the resistance of a platinum resistance thermometer (PRT). AB is a uniform resistance wire of length 1.0 m and C is a sliding contact on this wire. A standard resistor R is included in the circuit. The position of C is adjusted until the voltmeter V reads zero. Figure 1 (i) By applying Kirchhoff’s laws to loops ADCA and BCDB, deduce an expression for the resistance of the PRT in terms of l1, l and the value of the standard resistor. (5 marks) (ii) The PRT consists of 9.0 m of wire of diameter 8.0 !! 10?2 mm. The voltmeter reads 0 V when l1 = 0.44 m. If the standard resistor, R, has a resistance of 224 ?, what is the resistivity of platinum? Show that you have checked that the value for the resistivity and its unit are sensible. (9 marks) (c) Indicate briefly what factors might affect the precision of the measurement when using such a simple circuit to measure the resistance of the PRT. (2 marks)

Transcript

00:01 The given circuit has been redrawn as shown and ac is the part of the wire having length l1, let its resistance is r1.

00:18 Bc is the another part of the wire having length l minus l1, this is the length l minus l1 and its resistance is r2.

00:33 And l is the total length of the wire acp and the prt, the unknown resistance which we want to calculate is connected in the branch ad and a standard resistance capital r is connected in the branch db and rest is the same like a given circuit.

00:59 Let the total current flowing through the battery is i and let the means out of the current i the means current i1 flows in the branch ac in the branch ac now as per the kirch of law total incoming current should be equal to total outgoing current so means current in the branch ad should be equal to i minus i 1 so current in the branch ad is equal to i minus i 1 now no current flows through the branch cd say because the voltmeter shows zero deflection no current but still for the calculation purpose we call it a value of iv is is the current through the same means say means branch cd right so again we apply the kirchhoff means current law at point c total incoming means current is i1 and outgoing current in the branch cd is iv so outgoing means current in the branch c b should be equal to i1 minus iv similarly at point d the incoming current in the branch a d is i minus i i i 1 and incoming current in the branch in the branch cd is means iv so total well means current coming out of the branch d at point d the total incoming current a d plus cd should be equal to d b so this is the means current in the branch db right so now we apply two voltage loops and we apply the kirchof voltage law in this circuit and we have first loop a d c b it is a d c and b right this is the direction of the loop a d c b right and second loop is b c d b and this is the second loop b c d b now we will apply the means kitsch of voltage law in these loops so in the loop acdb we apply the means of voltage law in the loop adca.

03:46 Now in the loop adca we start say from a then voltage drop across pprt would have the negative sign because the direction of loop is from a to d and the direction of the current in the branch ad which is which is means i minus i1 is also from a to d so drop will be negative as per the kirch of the voltage law similarly in in the branch cd the means the direction of the loop is from d to c while the direction of the current is from c to d to drop will be positive the direction of current is say from a to c which is the direction of means i1 while the direction of loop is from c to a which is a opposite direction so the drop at the resistance r1 which is nothing but the resistance of the wire of length l1 this drop will be we say will be negative and drop we know the say from omslaw is equal to drop is equal to voltage into current so drop at pr t will be equal to minus of i minus means i1 into resistance of the pr which we want to calculate.

05:17 The same is for the remaining loops.

05:21 So we write the equations.

05:24 The first equation will be for loop adca, it is minus of i minus i1 into resistance of prt.

05:42 Let r prt is the resistance of the p r t say plus iv into r v is the in the branch cd rv is the resistance of the voltmeter and say plus i1 r1 is equal to zero so as there is the voltmeter it means there is no current in the cd branch central branch this is zero we put value this 0 and if we solve this it comes out to be i 1 r1 is equal to i minus i 1 into r pr t right this is equation number 1 same way we are we apply the equation in loop b c d b b b right here i 1 minus iv multiplied by r2 how let us check in the loop b c b b the direction of in the branch bc the direction of loop is from b to c while the means current is is is is is saying is flowing from c to b which is in the opposite direction that is why the drop in the part bc will be positive and this drop is equal to voltage.

07:26 Sorry, the drop is equal to current, which is i1 minus iv multiplied by the resistance.

07:36 So, and the remaining part also we write iv, which is for the central branch minus capital r for the unknown resistance.

07:47 Same means branch.

07:49 I1, say plus iv is equal to 0.

07:53 If we solve this further, it becomes i1.

07:58 R2 is equal to capital r i minus i1 this is equation 2 right this is i 1 r2 this is i 1 right we write here again i 1 r2 now we divide equation 1 by equation 2 divide equation 1 by equation 2 divide equation 1 by equation 2 we get r1 upon r2 is equal to r p r t resistance of the pr which we want to calculate divided by the unknown resistance now say from the basic formula of resistance we know the resistance is equal to resistivity length upon area right so on on this basis r1 is equal to to row l1 upon a let row is the resistivity of the wire it means ac b and a is the area of the cross section of the same wire and l1 is the length for the z zero deflection which is nothing but the length of part part ac and r2 is equal to row l minus l1 upon a right it is given in the problem that that part of the length of the part c b is is means l minus a is l1 if we solve this further this gives r1 upon r2 is equal to l1 upon l1 right so we put this value of r2 here this value of r1 by r2 we put here right so we get an expression l1 upon l minus l1 is equal to r pr t upon r pr t upon capital r further we solve the expression for the pr t which is resistance of the prt, rprt, is equal to capital r, l1 upon l minus l1.

11:05 This is the required expression.

11:08 The value of prt resistance in terms of the standard resistance capital r and the length of the zero reflection and zero deflection l1.

11:20 This is the answer for part one.

11:27 Now in part two, we are supposed to find out the resistivity of the platinum using this experiment and also we need to compare it with the standard value of the resistivity of the platinum.

11:45 Now in part two, here they say the standard, the prt resistance has a length.

12:02 Of 9 meter its diameter is 8 mm means its resistance is 4 mm which is 8 by 2 right and the length of zero deflection of volt meter l1 is equal to 0 .44 meter right so it is given in the problem the total length is 1 meter so l minus l1 should be equal to 1 minus 0 .44 which is equal to 0 .56 meter.

12:44 And the standard resistance r has a value of 224 oms.

12:54 These all details are given...

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