As per the given question,
Current in the appliances (i)=10×106A(i)=10\times 10^6A(i)=10×106A
Resistance of the appliances (R)=8×103Ω(R)=8\times 10^3 \Omega(R)=8×103Ω
We know that, power dissipation in the appliances (P)=i2R(P)=i^2R(P)=i2R
P=(10×106)2×8000=8×1017wattP=(10\times 10^6)^2\times8000=8\times 10^{17}wattP=(10×106)2×8000=8×1017watt
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