Question 56

2a. Use $R=\frac{\rho L}{A}$ to give $R=3.7\Omega$ which is the same as $37m\Omega km^{-1}$ because the line is $100km$ long.

2b. To do this question, it helps to have an idea of what is actually happening to the current. It leaves the power station along one of the cables, travels through some kind of network which we don't need to know about then returns along the other cable. Working out this current is crucial to this question and can be found using $P=IV$. Once you have this current, use $P=I^2R$ to find the power dissipated in each cable and multiply by 2 because there are two of them. Alternatively, you can use $V=IR$ to work out the voltage dropped across each cable and then use $P=\frac{V^2}{R}$. Don't be tempted to use the $V$ from the question in $P=\frac{V^2}{R}$ because 400kV is the voltage dropped across the cables $and$ the city, not just the cables.

$P_{total}=IV$, $I=\frac{10^9}{400\times10^3}=2500A$

$P_{cables}=I^2R_{cables}=2500^2\times2\times3.67=46MW$

2c. $1KWh=1000\times3600=3.6\times10^6J$

$1GWh=10^9\times3600=3.6\times10^{12}J$

2d. Answers A,B,D and E are simple to eliminate as follows:

A: The efficiency of the power station is of no consequence after you quote its power output which would have already had the efficiency taken into account if necessary.

B: Similarly, the appliances consume a certain power (as quoted in the question) and still use this power whether they are efficient or not.

D. This assumes that the energy can't be lost in wasted forms.

E. This also assumes that the energy can't be lost in wasted forms.

2a. Use $R=\frac{\rho L}{A}$ to give $R=3.7\Omega$ which is the same as $37m\Omega km^{-1}$ because the line is $100km$ long.

2b. To do this question, it helps to have an idea of what is actually happening to the current. It leaves the power station along one of the cables, travels through some kind of network which we don't need to know about then returns along the other cable. Working out this current is crucial to this question and can be found using $P=IV$. Once you have this current, use $P=I^2R$ to find the power dissipated in each cable and multiply by 2 because there are two of them. Alternatively, you can use $V=IR$ to work out the voltage dropped across each cable and then use $P=\frac{V^2}{R}$. Don't be tempted to use the $V$ from the question in $P=\frac{V^2}{R}$ because 400kV is the voltage dropped across the cables $and$ the city, not just the cables.

$P_{total}=IV$, $I=\frac{10^9}{400\times10^3}=2500A$

$P_{cables}=I^2R_{cables}=2500^2\times2\times3.67=46MW$

2c. $1KWh=1000\times3600=3.6\times10^6J$

$1GWh=10^9\times3600=3.6\times10^{12}J$

2d. Answers A,B,D and E are simple to eliminate as follows:

A: The efficiency of the power station is of no consequence after you quote its power output which would have already had the efficiency taken into account if necessary.

B: Similarly, the appliances consume a certain power (as quoted in the question) and still use this power whether they are efficient or not.

D. This assumes that the energy can't be lost in wasted forms.

E. This also assumes that the energy can't be lost in wasted forms.