circuit solving

The Wheatstone Bridge gives 2 parallel lines of 16 ohms. NET = 8 OHM.

The current flows through the short circuited diagonal wire. NET resistance= 4+8 = 12 in parallel with 8 ohm.

NET : 4X8/(4+8) =4.8

THIS is in series with 5 ohm. SO Rnet=5+4.8= 9.8

• the wheatstone bridge is balanced therefore the 4ohm resistor can be disregarded; the equivalent resistance is 8 ohms
• the 7 + 1 resistors are short circuited
• equivalent is 5 + (8 // 12) = 5 + 4,8 = 9,8 ohms.

From the Wheatstone Bridge, net resistance 16||16=8 ohm. Current wont folw through 7 ohm & 1 ohm because'f that short circuit. so, in total, resistance 5+(4+8)||(8)=9.8 ohm

Well this was an easy one, first of all the circuit element having 1Ω and 3Ω resistors in series is short-circuited by the wire parallel to it, also, the 4Ω resistor in between the two 8Ω-8Ω resistors is removed due to the no current flow (Wheatstone Bridge). So, the circuit solves to be a series-parallel circuit comprising of 5Ω in series with 3 parallel circuit element [two 8Ω-8Ω resistors (in series) and one 4Ω-8Ω resistors (in series)].

R = 5 + 1/[1/(8+8)+1/(8+8)+1/(4+8)] = 9.8

wheatstone bridge (8,8,8,8,4)resistors and short circuit at 7 and 1 ohm resistors.