Question

resistance of 500 ohm and 3000 ohm are placed in series with a 60 V supply. What will be the reading on a voltmeter of internal resistance of 2000 ohm when placed across-
1) 500 ohm resistors
2) 3000 ohm resistors

Answer 1

The resistance of 500 ohm and 3000 ohm are placed in series with a $60\mathrm{V}$ supply. What will be the reading on a voltmeter of internal resistance of 2000 ohm when placed across-

1) 500 ohm resistors;

2) 3000 ohm resistors.

Solution.

$$R _ {1} = 5 0 0 O h m, R _ {2} = 3 0 0 0 O h m, V = 6 0 V, R _ {V} = 2 0 0 0 O h m;$$ $$V _ {1} -? V _ {2} -?$$

1)

The resistance between the points $A$ and $B$ :

$$\frac {1}{R _ {A B}} = \frac {1}{R _ {1}} + \frac {1}{R _ {V}};$$

$R_{V}$ - the internal resistance of the voltmeter.

$$R _ {A B} = \frac {R _ {1} R _ {V}}{R _ {1} + R _ {V}}.$$

The resistance between the points $A$ and $C$ :

$$R _ {A C} = R _ {A B} + R _ {2};$$ $$R _ {A C} = \frac {R _ {1} R _ {V}}{R _ {1} + R _ {V}} + R _ {2};$$ $$R _ {A C} = \frac {R _ {1} R _ {V} + R _ {1} R _ {2} + R _ {V} R _ {2}}{R _ {1} + R _ {V}}.$$

By Ohm's law the amperage $I_{1}$ is:

$$I _ {1} = \frac {V}{R _ {A C}};$$ $$I _ {1} = \frac {V (R _ {1} + R _ {V})}{R _ {1} R _ {V} + R _ {1} R _ {2} + R _ {V} R _ {2}}.$$

The voltage across resistance $R_{1}$ is the same as the voltage across voltmeter and the same as the voltage across the points $A$ and $B$ then:

$$V _ {1} = I _ {1} R _ {A B};$$ $$V _ {1} = \frac {V (R _ {1} + R _ {V})}{R _ {1} R _ {V} + R _ {1} R _ {2} + R _ {V} R _ {2}} \cdot \frac {R _ {1} R _ {V}}{R _ {1} + R _ {V}};$$ $$V _ {1} = \frac {V R _ {1} R _ {V}}{R _ {1} R _ {V} + R _ {1} R _ {2} + R _ {V} R _ {2}}.$$ $$V _ {1} = \frac {6 0 \cdot 5 0 0 \cdot 2 0 0 0}{5 0 0 \cdot 2 0 0 0 + 5 0 0 \cdot 3 0 0 0 + 2 0 0 0 \cdot 3 0 0 0} = 7. 0 6 (V).$$

2)

The resistance between the points $B$ and $C$ :

$$\frac {1}{R _ {B C}} = \frac {1}{R _ {2}} + \frac {1}{R _ {V}};$$

$R_{V}$ - the internal resistance of the voltmeter.

$$R _ {B C} = \frac {R _ {2} R _ {V}}{R _ {2} + R _ {V}}.$$

The resistance between the points $A$ and $C$ :

$$R_{AC} = R_1 + R_{BC};$$ $$R_{AC} = R_1 + \frac{R_2 R_V}{R_2 + R_V};$$ $$R_{AC} = \frac{R_1 R_2 + R_1 R_V + R_2 R_V}{R_2 + R_V}.$$

By Ohm's law the amperage $I_2$ is:

$$I_2 = \frac{V}{R_{AC}};$$ $$I_2 = \frac{V(R_2 + R_V)}{R_1 R_2 + R_1 R_V + R_2 R_V}.$$

The voltage across resistance $R_2$ is the same as the voltage across voltmeter and the same as the voltage across the points $B$ and $C$ then:

$$V_2 = I_2 R_{BC};$$ $$V_2 = \frac{V(R_2 + R_V)}{R_1 R_2 + R_1 R_V + R_2 R_V} \cdot \frac{R_2 R_V}{R_2 + R_V};$$ $$V_2 = \frac{V R_2 R_V}{R_1 R_V + R_1 R_2 + R_V R_2}.$$ $$V_2 = \frac{60 \cdot 3000 \cdot 2000}{500 \cdot 2000 + 500 \cdot 3000 + 2000 \cdot 3000} = 42.35(V).$$

Answer:

1) $V_1 = 7.06V$;

2) $V_2 = 42.35V$.