Question

Two 20 kiloOhms resistors (R1 and R2) are connected in series across a 100 volt supply. Calculate:

(a) The Voltage drop across the R1 resistor from nominal values:

(b) The voltage across the R1 resistor if measured using a digital voltmeter (internal resistance of 1 milliOhms)

(c) The voltage across the R1 resistor if measured using an analogue voltmeter (internal resistance of 20 kiloOhms)

Answer 1

Two 20 kiloOhms resistors (R1 and R2) are connected in series across a 100 volt supply. Calculate:

(a) The Voltage drop across the R1 resistor from nominal values:

(b) The voltage across the R1 resistor if measured using a digital voltmeter (internal resistance of 1 milliOhms)

(c) The voltage across the R1 resistor if measured using an analogue voltmeter (internal resistance of 20 kiloOhms)

Solution

a) $U_{1} = IR_{1} = \frac{U}{R_{1} + R_{1}} R_{1} = \frac{U}{2} = 50\mathrm{V}$

b) $U = I * R + I_{1} * R_{1}$

$$\begin{array}{l} I = \frac {U}{R _ {1} + \frac {R _ {1} r}{R _ {1} + r}} = \frac {1 0 0}{2 0 0 0 0 + \frac {2 0 0 0 0 * 0 . 0 0 1}{2 0 0 0 0 + 0 . 0 0 1}} = 0, 0 0 5 A \\ \left\{ \begin{array}{l} I _ {1} R _ {1} = I _ {2} r \\ I = I _ {1} + I _ {2} \end{array} \right. \gg I _ {1} = I * \frac {r}{R _ {1} + r} = 0, 0 0 5 * \frac {0 . 0 0 1}{2 0 0 0 0 + 0 . 0 0 1} = 2, 5 * 1 0 ^ {- 1 0} A \\ U _ {1} = I _ {1} R _ {1} = 2, 5 * 1 0 ^ {- 1 0} * 2 0 0 0 0 = 5 * 1 0 ^ {- 6} V \\ \end{array}$$

c) $I = \frac{U}{R_1 + \frac{R_1r}{R_1 + r}} = \frac{100}{20000 + \frac{20000*20000}{20000 + 20000}} = 0,0033A$

$$\begin{array}{l} I _ {1} = I * \frac {r}{R _ {1} + r} = 0, 0 0 5 * \frac {2 0 0 0 0}{2 0 0 0 0 + 2 0 0 0 0} = 0, 0 0 2 5 A \\ U _ {1} = I _ {1} R _ {1} = 0, 0 0 2 5 * 2 0 0 0 0 = 5 0 V \\ \end{array}$$