The aim of this Apprentice Corner article is to consider how voltages, currents and power are affected by series-connected loads.
Introduction
The resistance of a material is dependent upon conductor material (resistivity), length, cross-sectional and temperature. Within this article it will be shown that the way in which loads are connected can affect the current drawn from the supply, the voltage lost due to circuit resistance and the power available at the load.Only resistive loads will be considered. The effects of reactance will be dealt with in a subsequent article.
It would be useful if you had to hand the Site Guide produced by Certsure when dealing with resistivity and temperature coefficients considered in this article.
Series connected resistors
For series connected resistors as shown in Fig 1, the total resistance is the sum of each individual resistor: Rt = R1 + R2 + R3 + ….Rn.
The current drawn from the supply will pass through each resistor equally and will therefore be a constant value (always remember: current is constant in a series circuit1). There will be a voltage drop across each resistor that, when added together, will equal the supply voltage2.
- Kirchhoff ’s current law – the algebraic sum of the currents at a junction in a network is zero.
- Kirchhoff ’s voltage law – in a closed loop, the algebraic sum of the emfs acting round the loop is equal to the algebraic sum of the potential difference in the loop.
Voltage drop across a resistor: U1 = I × R1, U2 = I × R2 and U3 = I × R3
The power consumed by each resistor in a purely resistive circuit can be found by one of three methods:
- Method 1, using the supply current and the resistance: P = I2 × R
- Method 2, using the supply current and the voltage dropped across a resistor: P = U × I
- Method 3, using the voltage dropped across a resistor and the resistance:
Since the current is constant in a series circuit, the voltage is divided across the number of resistive elements in the circuit. The voltage dropped across a resistor in a series circuit is equal to the value of that resistor divided by the total circuit resistance and multiplied by the supply voltage.
For example, the voltage across R1 can be found from:
where US is the supply voltage.
It is useful to get into the habit of using subscripts to denote the element being determined, as it makes it easy to follow the working out process.
Example
A radial circuit wired in PVC/PVC flat twin and earth cable fed from a 230 V supply has three resistive loads, connected as shown in Fig 2. What is the volt-drop in each section and the voltage at each load?
The resistance of the circuit cable can be found in one of two ways. Table B1 of Appendix B of the Site Guide (published by Certsure) shows that 6mm2 copper conductor has a resistance of 3.08 mΩ/m (per conductor line or neutral) at 20 °C.
Alternatively, from table 4D5 of Appendix 4 of BS 7671, a 6 mm2 cable has a mV/A/m value of 7.3 at 70 °C (in effect mΩ/m). When either value is adjusted to the other temperature, the values are almost identical.
For example, converting 7.3 mV/A/m from 70 °C to 20 °C:
7.3 = R20 [1 + 0.004 (70 – 20)] and transposing gives
This compares very closely with the value from Table B1 of the Site Guide: 2 × 3.08 = 6.16 mΩ per metre:
Volt-drop in section O – A:
UOA = I × R = 42 × (2 × 3.08 ×10-3 × 15) = 3.88 V
Voltage at A:
UA = U – UOA = 230 – 3.88 = 226.12 V
Volt-drop in section A – B:
UAC = I × R = 22 × (2 × 7.41 ×10-3 × 12) = 3.91 V
Voltage at B:
UB = UA – UAB = 226.12 – 3.91 = 222.21 V
Volt-drop in section B – C:
UBC = I × R = 10 × (2 × 12.1 × 10-3 × 5) = 1.21 V
Voltage at C:
UC = UB – UBC = 222.21 – 1.21 = 221 V
Total voltage dropped due to cable resistance is 9 V.
From Table 4Ab of Appendix 4 of BS 7671 the maximum allowed voltage drop is 5% (11.5 V) for a radial circuit other than a lighting circuit, therefore this circuit satisfies the volt-drop constraints.
Scenario 1
A customer complains that their electric shower doesn’t heat the water sufficiently. The shower nameplate says the shower is rated at 8 kW at 240 V. The shower circuit has a length of run 26 m in total and is wired using 6mm2 copper conductors and installed to reference method C. The supply voltage is measured to be 223 V. What is the actual power the shower is able to deliver?
The circuit arrangement is shown in Fig 3 and the simplified circuit is shown in Fig 4.
To calculate the voltage at the shower terminals would normally involve determining the design current (Ib) and using that in the volt-drop equation (mV/A/m × Ib × l × 10-3). However, in this instance this method is not ideal since the shower rating is at a non-standard voltage. A different approach is required, which is to consider the shower forming a series circuit and to use Ohm’s law.
Using the resistance per metre values from Table B1 of Appendix B of the Site Guide:
Cable resistance from supply to isolator: R1 = 2 × 3.08 × 10-3 × 20 = 0.1232 Ω
Cable resistance from isolator to shower: R2 = 2 × 3.08 × 10-3 × 6 = 0.037 Ω
Resistance of shower element:
Voltage at the shower terminals:
It was mentioned earlier that power is proportional to the voltage squared therefore, power available from the shower:
The voltage at the supply and hence at the load appears low in value, but they are within the tolerance permitted by clause 27(3) of the Electricity Safety, Quality and Continuity Regulations, which is 230 V – 6% +10%, giving a range of 216.2 V and 253 V.
A future article will consider loads connected in parallel.
Multiple-choice questions
1. Four resistors of value 46 Ω, 54 Ω, 66 Ω and 88 Ω are connected in series. It is required to modify their combined resistance to 280 Ω by replacing one of the existing resistors with a new resistor of value 80 Ω. Which original resistor should be replaced?a) 46 Ω
b) 54 Ω
c) 66 Ω
d) 88 Ω
2. The element that is constant in a series circuit is:
a) Current
b) Voltage
c) Resistance
d) Power
3. The current through a heating element is 6 A when the voltage is 120 V. What voltage must be applied to obtain a current of 4 A?
a) 240 V
b) 120 V
c) 180 V
d) 80 V
4. The current flow through the field coils of a motor is 4 A when the resistance is 500 Ω. Due to a rise in temperature, the resistance increases to 550 Ω. If the voltage remains the same, what will be the new current value?
a) 3.64 A
b) 4.4 A
c) 2 A
d) 4 A
5. In the scenario considered earlier, what will be the power available at the shower if the supply voltage was increased to 253 V (230 V +10%)?
a) 8.0 kW, cannot exceed maximum rating of shower
b) 8.5 kW, power increased due to voltage increase
c) 6.6 kW, current is constant in a series circuit
d) 7.0 kW, due to increased power loss in the circuit
6. For the series circuit of Fig MC1, determine the value of R1
a) 6 Ω
b) 2 Ω
c) 3 Ω
d) 1 Ω
Apprentice Corner answers
1. Correct option is (b)Total resistance: Rt = R1 + R2 + R3 + R4 = 46 + 54 + 66 + 88 = 254 Ω
If the new value is to be 280 Ω, the resistor to replace for the 80 Ω one is the 54 Ω.
2. Correct option is (a)
Current is constant in a series circuit.
3. Correct option is (d)
The resistance of the heater element is a constant.
A smaller current will need a reduced voltage.
4. Correct option is (a)
The supply voltage is constant.
Therefore: U = I1 × R1 and U = I2 × R2 ∴ I1 × R1 = I2 × R2
If the resistance increases, the current will reduce.
5. Correct option is (b)
The power available at the shower:
Whilst a shower is a type of load that will not produce an overload current due to having a fixed heater resistance, that is not to say that the load current is fixed. The current flow is dependent upon the supply voltage.
6. Correct option is (c)
