V=K(h)a(I)b(G)c(C)d (V is voltage)
We know [h]=ML2T−1
[I]=A
[G]=M−1L3T−2
[C]=LT−1
[V]=ML2T−3A−1
ML2T−3A−1=(ML2T−1)a(A)b(M−1L3T−2)c(LT−1)d
ML2T−3A−1=Ma−cL2a+3c+dT−a−2c−dAb
a−c=1.....(1)
2a+3c+d=2.....(2)
−a−2c−d=−3.....(3)
b=−1...(4)
On solving
c=−1
a=0
d=5,b=−1
V=K(h)0(A)−1(G)−1(C)5