The circuit shown in Fig-I is replaced by that in FigII.

The circuit shown in Fig-I is replaced by that in FigII. If current I remains the same, then R₀ will be

From Fig. (a),

I =	V	+	V	+	V	=	7V
	32R		16R		8R		32R

From Fig. (b),

I = (V - IR₀)		1	+	1	+	1		=	7(V- IR₀)
		4R		2R		R			4R

Since current remains same

∴	7(V- IR₀)	=	7V
	4R		32R

⇒ (V- IR₀) =	V
	8

⇒ IR₀ =	7V	= 4R.I
	8

⇒ R₀ = 4R