Calculation of additional resistance

Concepts and formulas

If the consumer must be turned on at a higher voltage than what it is designed for, they are turned on in series with it additional resistance rd (Fig. 1). The additional resistance creates voltage drop Ud, which reduces the user's voltage to the required value Up.

The source voltage is equal to the sum of the consumer voltages and the additional resistance: U = Up + Ud; U = Upn + I ∙ rd.

From this equation it is possible to determine the required additional resistance: I ∙ rd = U-Up, rd = (U-Up) / I.

Reducing the voltage using an additional resistance is uneconomical, because in the resistance the electrical energy is converted into heat.

Rice. 1. Additional resistance

Examples of

1. An arc lamp (Fig. 2) consumes a current I = 4 A at an arc voltage Ul = 45 V. What resistance should be connected in series with the lamp if the DC supply voltage is U = 110 V?

Rice. 2.

In fig.2 shows a diagram of the inclusion of graphite electrodes and additional resistance, as well as a simplified diagram with the designation of the resistance and the arc lamp.

The current I = 4 A passing through the lamp and the additional resistance rd will create a useful voltage drop across the arc Ul = 45 V, and through the additional resistance a voltage drop Ud = U-Ul = 110-45 = 65 V.

Additional resistance rd = (U-Ul) / I = (110-45) / 4 = 65/4 = 16.25 Ohm.

2. A mercury lamp with an operating voltage of 140 V and a current of 2 A is connected to a 220 V network through an additional resistance, the value of which must be calculated (Fig. 3).

Rice. 3.

The voltage in the network is equal to the sum of the voltage drop in the additional resistance and in the mercury lamp:

U = Ud + Ul;

220 = I ∙ rd + 140;

2 ∙ rd = 220-140 = 80;

rd = 80/2 = 40 ohms.

With additional resistance, voltage drops only when current flows through it. When it is switched on, the full mains voltage drops to the lamp, since in this case the current is small. The current and voltage drop across the additional resistance increase gradually.

3. A 40 W gas discharge lamp with an operating voltage of 105 V and a current of 0.4 A is connected to a 220 V network. Calculate the value of the additional resistance rd (Fig. 4).

The additional resistance must reduce the mains voltage U to the operating voltage of the lamp Ul.

Rice. 4.

To light the lamp, a mains voltage of 220 V is first required.

U = Ud + Ul;

Ud = 220-105 = 115 V;

rd = (115 V) / (0.4 A) = 287.5 Ohm.

The voltage drop across the resistance results in a loss of electrical energy, which is converted into heat.In alternating current, a choke is used instead of an additional resistance, which is much more economical.

4. A vacuum cleaner designed for voltage Uc = 110 V and power 170 W must operate at U = 220 V. What must be the additional resistance?

In fig. 5 shows a sketch and schematic diagram of a vacuum cleaner, showing the motor D with fan and additional resistance.

The supply voltage is divided between the motor and the additional resistance rd in half so that the motor has 110V.

U = Udv + Ud;

U = Udv + I ∙ rd;

220 = 110 + I ∙ rd.

We calculate the current according to the data of the vacuum cleaner:

I = P / Us = 170/110 = 1.545 A;

rd = (U-Udv) / I = (220-110) / 1.545 = 110 / 1.545 = 71.2 Ohm.

Rice. 5.

5. The DC motor for a voltage of 220 V and a current of 12 A has internal resistance rv = 0.2 ohms. What should the resistance be? starting rheostatso that the inrush current at start-up is not more than 18 A (Fig. 6)?

Rice. 6.

If you connect the motor directly to the network, without starting resistance, then the starting current of the motor will have an unacceptable value Iv = U / rv = 220 / 0.2 = 1100 A.

Therefore, to turn on the motor, it is necessary to reduce this current to approximately I = 1.5 ∙ In. During normal operation of the motor, the rheostat is short-circuited (the motor is in position 5), since the motor itself creates a voltage directed against mains voltage; therefore, the nominal motor current has a relatively small value (In = 12 A).

When starting, the current is limited only by the starting rheostat and the internal resistance of the motor: I = U / (rd + rv);

18 = 220 / (rd + 0.2); rd = 220 / 18-0.2 = 12.02 Ohm.

6.The voltmeter has a measuring range of Uv = 10 V and its resistance rv = 100 Ohm. What should be the additional resistance rd in order for the voltmeter to measure voltages up to 250 V (Fig. 7)?

Rice. 7.

The measuring range of the voltmeter is increased when the series additional resistance is included. The measured voltage U is divided into two voltages: the voltage drop across the resistance Ud and the voltage at the terminals of the voltmeter Uv (Fig. 8):

Rice. eight.

U = Ud + Uv;

250 V = Ud + 10 B.

The current passing through the device, with full deflection of the arrow, will be equal to: Iv = Uv / rv = 10/100 = 0.1 A.

The same current should pass through the voltmeter when measuring a voltage of 250 V (with an additional resistance included).

Then 250 B = Ic ∙ rd + 10 B;

Iv ∙ rd = 250-10 = 240V.

Additional resistance rd = 240 / 0.1 = 2400 Ohm.

With any additional resistance, the deflection of the voltmeter needle will be maximum when the voltmeter voltage is 10 V, but its scale is calibrated according to the additional resistance.

In our case, the maximum deviation of the arrow should correspond to a division of 250 V.

In general, the range gain of the voltmeter will be:

n = U / Uv, or n = (Ud + Uv) / Uv = Ud / Uv +1;

n-1 = (Ic ∙ rd) / (Ic ∙ rc);

rv ∙ (n-1) = rd;

rd = (n-1) ∙ rv.

7. The internal resistance of the voltmeter is 80 Ohm with a measuring range of 30 V. Calculate the required value of the additional resistance rd so that the voltmeter can measure a voltage of 360 V.

According to the formula obtained in the previous calculation, the additional resistance is: rd = (n-1) ∙ rv,

where the range gain is n = 360/30 = 12.

Therefore,

rd = (12-1) ∙ 80 = 880 Ohms.

The additional resistance rd for the new 360 V measurement range will be 880 Ohm.