Question

Anyone help me with the 45 th problem plzz...make it fast...

Answer 1

The equivalent resistance for the infinite network is given as,

R′=2+R′R′+1

Where, the equivalent resistance for the infinite network is R′.

By simplifying the above expression, we get

(R′)2−2R′−2=0R′=2±4+8√2R′=1±3–√

Since, the resistance cannot have negative value therefore,

R′=1+3–√=2.73 Ω

The current drawn from the supply is given as,

I=Vr+R′

Where, the supply voltage is V and the internal resistance is r.

By substituting the given values in the above expression, we get

I=120.5+2.73=123.23=3.72 A

Thus, the current drawn from the supply is 3.72 A

Hope helps

Answer 2

hey mate:

Your answer:-

Solution :

Let the equivalent resistance of the network be x×1x×1

Rs=1+1+1+1Rs=1+1+1+1

x=xΩx=xΩ

The effective resistance x and 1Ω1Ω are in parallel

∴1Rp=1x+11∴1Rp=1x+11

=1+xx=1+xx

Rp=x1+xRp=x1+xΩΩ

Now, the resistance Rp,1ΩRp,1Ω and 1Ω1Ω are in series.

∴R=Rp+1+1=x1+x∴R=Rp+1+1=x1+x+1+1+1+1

=x1+x=x1+x+2+2

For infinite resistance

x=x1+xx=x1+x+2+2

(ie) x(1+x)=x+2x(1+x)=x+2

=> x2−2x−2=0x2−2x−2=0

applying the quadratic formula

1±4+8−−−−√21±4+82

=1±3–√=1±3

neglecting the negative value,

Resistance of the network

x=1+3–√x=1+3

=1+1.732=1+1.732

=2.732Ω=2.732Ω

Total resistance of the circuit =2.732+0.5=3.232Ω=2.732+0.5=3.232Ω

Current I=V3.232=123.232I=V3.232=123.232

=3.72A=3.72A

Answer : 3.72A