Question

problem 6: (6% of assignment value) a 240 kv power transmission line carrying 5.00×10² a is hung from grounded metal towers by ceramic insulators, each having a 1.5×10⁹ ω resistance.
part (a) what is the resistance, rp, in ohms, to the ground of 100 of these insulators along the line? rp = 1.5×10⁷ correct!
part (b) calculate the power, p, in watts, that is dissipated by 100 of these insulators. p =

Explanation:

Step1: Recall power - dissipation formula

The power dissipated in a resistor is given by $P = I^{2}R$. Here, we know the resistance of one insulator $R_p=1.5\times 10^{9}\ \Omega$ and the current $I = 5.00\times 10^{2}\ A$.

Step2: Calculate power dissipated in one insulator

Using the formula $P_1=I^{2}R_p$, substitute $I = 5.00\times 10^{2}\ A$ and $R_p=1.5\times 10^{9}\ \Omega$. So, $P_1=(5.00\times 10^{2})^{2}\times1.5\times 10^{9}$. First, $(5.00\times 10^{2})^{2}=5.00^{2}\times10^{4}=25\times 10^{4}=2.5\times 10^{5}$. Then $P_1 = 2.5\times 10^{5}\times1.5\times 10^{9}=3.75\times 10^{14}\ W$.

Step3: Calculate power dissipated in 100 insulators

Since we have 100 insulators, the total power $P = 100\times P_1$. So $P=100\times3.75\times 10^{14}=3.75\times 10^{16}\ W$.

Answer:

$3.75\times 10^{16}$