Figure 11
(a) Electricity supplied to homes has a frequency of
☐ A 0.02 Hz
☐ B 20 Hz
☐ C 50 Hz
☐ D 500 Hz
(b) Explain why the National Grid uses high voltages with small currents to transfer electricity from power stations - Edexcel - GCSE Physics - Question 5 - 2023 - Paper 2
Question 5
Figure 11
(a) Electricity supplied to homes has a frequency of
☐ A 0.02 Hz
☐ B 20 Hz
☐ C 50 Hz
☐ D 500 Hz
(b) Explain why the National Grid uses high voltages wit... show full transcript
Worked Solution & Example Answer:Figure 11
(a) Electricity supplied to homes has a frequency of
☐ A 0.02 Hz
☐ B 20 Hz
☐ C 50 Hz
☐ D 500 Hz
(b) Explain why the National Grid uses high voltages with small currents to transfer electricity from power stations - Edexcel - GCSE Physics - Question 5 - 2023 - Paper 2
Step 1
Electricity supplied to homes has a frequency of
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The frequency of electricity supplied to homes is typically 50 Hz. Therefore, the correct answer is C 50 Hz.
Step 2
Explain why the National Grid uses high voltages with small currents to transfer electricity from power stations.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The National Grid uses high voltages for electricity transmission in order to minimize energy loss over long distances. Higher voltage enables the use of lower currents, which reduces the energy lost as heat due to the resistance in the wires. This method is efficient for transporting large amounts of electrical energy from power stations to consumers.
Step 3
Calculate the power in the primary coil.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To calculate the power in the primary coil, we use the formula:
P=V×I
where P is power, V is voltage, and I is current. To provide a specific numerical answer, the values for voltage and current must be known. Assuming some values:
If V = 200 V and I = 0.1 A:
P=200extV×0.1extA=20extW
Thus, power in the primary coil = 20 W.
Step 4
For the transformer in Figure 12, evaluate, in its simplest form, the ratio secondary voltage : primary voltage
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To evaluate the ratio of secondary voltage to primary voltage for the transformer, we must know the number of turns in the primary and secondary coils. Assuming the turns ratio is represented as:
rac{V_s}{V_p} = rac{N_s}{N_p}
where Vs is the secondary voltage, Vp is the primary voltage, Ns is the number of turns in the secondary coil, and Np is the number of turns in the primary coil. Without specific values, we cannot compute the exact ratio.