(a) Redraw the circuit to show how a voltmeter would be connected to measure the voltage across

(a) Redraw the circuit to show how a voltmeter would be...

Key Concepts

Voltage Divider Principle

The voltage divider principle is used to determine the voltage across a component in a series circuit. It explains how the applied voltage is divided among the resistors in proportion to their resistances, which is critical when calculating voltage drops in a circuit.

Ideal Voltmeter

An ideal voltmeter is a theoretical measurement device that has infinite resistance, meaning it does not draw any current from the circuit under test. This ensures that the circuit's behavior remains unchanged when the voltage is being measured.

Non-Ideal Voltmeter and Loading Effect

A non-ideal voltmeter has a finite (although often high) internal resistance, which means it draws a small amount of current from the circuit. This introduces a loading effect, slightly altering the circuit conditions and consequently the voltage reading, which must be considered in accurate measurements.

Parallel Connection for Voltage Measurement

When measuring the voltage across a component, the voltmeter should be connected in parallel with that component. This configuration allows the voltmeter to accurately capture the potential difference without significantly affecting the circuit, provided the meter’s resistance is sufficiently high.

Transcript

00:01 Calculate the voltage of resistance r4 by connecting what matter in the proper way to this resistance.

00:10 So first, in equation a, we have to redraw the circuit with the connected watt meter.

00:15 Let's do this.

00:16 So here is the battery, and it doesn't change.

00:21 Battery is connected in theory to r1, and r1 is connected in series to r2.

00:37 In parallel with r3 and r4.

00:53 And r4 here has to be measured.

00:56 Therefore, to measure it, we will connect voltmeter in parallel to r4.

01:07 So here is the answer to the first question.

01:09 Now, let's calculate the reading of the fault meter.

01:14 In order to do this, we first have to calculate the overall circuit resistance and the overall circuit resistance equals to r1.

01:24 R1 plus r234 where r234 is a resistance of last of the three resistances let's calculate r234 so r2 3 4 represents r2 connected in parallel with r3 in series with r4 so therefore reciprocal resistance of this part of the circuit equals to the sum of reciprocal 1 over r2 plus 1 over r34.

02:11 R34 equals to the sum of r3 and r4 because they are connected in cv.

02:18 So therefore the expression above can be changed a little bit.

02:30 And here we can calculate r34 immediately.

02:34 So it equals to the sum of r3 and r4, which is 16 kilo -ooms.

02:41 Plus 83 kilooms.

02:46 It equals to 99 kilooms.

02:54 So now we can calculate r2 -34.

02:58 It equals to r2 multiplied by r3 -4 divided by the sum.

03:08 And we can calculate it right away.

03:38 So now i need to use a calculator to get the number.

04:05 It equals to 1 .38.

04:09 Kilooms so and it means that the overall resistance equals to the sum of r1 which is 35 oms and r 234 which is 1 .38 kilooms so let's connect let's calculate it it equals to 1 .42 kilooms so this is a total resistance of the system now we can calculate the current flowing through the battery this current equals to emf of the battery divided by the total resistance or 9 .00 volts divided by 1 .42 kilooms it it equals to 6 .338 times 10 r by negative 3 ampires.

06:00 So this is a current flowing through the system through the emf.

06:09 So at this note, the current will be splited between r2 and r34, and we have to calculate the current flowing through r4.

06:20 So this current is a sum of i2 and i34 and the ratio of i34 over i2 is reciprocal to the ratios of the resistance is because the voltage is the same so it's r2 over r34 which is 1 .4 kilo r34 which is 99 kilombs.

07:27 So it equals to the following.

07:42 So it means that i -34 equals to on tiny fraction of i -2.

07:56 So now we can calculate i -2 and it equals to total current flowing through the battery.

08:42 Reduce the number of decimals so we can substitute the last one with four because we don't need that many decimals.

08:55 So now we can calculate i2.

08:58 It equals to 6 .34 times 10 negative 3 ampers over 1 .01 for 1.

09:12 And therefore i 3 -4 equals to the fraction of this current i2...

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