Help needed on wheatstone bridge problem - Electrical Engineering Stack Exchange most recent 30 from electronics.stackexchange.com 2019-08-26T07:38:39Z https://electronics.stackexchange.com/feeds/question/314200 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://electronics.stackexchange.com/q/314200 1 Help needed on wheatstone bridge problem BillyBob1 https://electronics.stackexchange.com/users/153770 2017-07-03T02:24:04Z 2017-07-03T02:44:29Z <p>My circuits 1 professor gave us some practice problems for an upcoming exam, and I'm not sure that I know how to do this one. <a href="https://i.stack.imgur.com/FCtYp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FCtYp.png" alt="enter image description here"></a></p> <p>My approach was to apply the circuit divider formula for the left and right sides, and that gave me the current across both parallel branches of the bridge, but I'm unsure of how to proceed. I was also wondering if I can place ground at some point on this circuit? Or would I need a voltage source to be able to do so? Thanks in advance.</p> https://electronics.stackexchange.com/questions/314200/-/314203#314203 0 Answer by sstobbe for Help needed on wheatstone bridge problem sstobbe https://electronics.stackexchange.com/users/148888 2017-07-03T02:43:28Z 2017-07-03T02:43:28Z <p>Your approach is correct. After finding the left-hand and right-hand currents \$I_{left}\$ and \$I_{right}\$. You apply KVL and write a loop around Vo, R2, and R4 as,</p> <p>$$+V_o + I_{right}R_4 - I_{left}R_2 = 0$$</p> <p>Rearrange for \$V_o\$</p> https://electronics.stackexchange.com/questions/314200/-/314204#314204 3 Answer by compumike for Help needed on wheatstone bridge problem compumike https://electronics.stackexchange.com/users/8398 2017-07-03T02:44:29Z 2017-07-03T02:44:29Z <p>R1 and R2 in series total 3K, while R3 and R4 in series total 6K, so the current will split 2/3rds to R1+R2, and 1/3rd to R3+R4. This is a fairly straightforward application of <a href="https://www.circuitlab.com/textbook/resistors-series-parallel/" rel="nofollow noreferrer">resistors in series and parallel</a>.</p> <p>You can choose a <a href="https://www.circuitlab.com/textbook/ground/" rel="nofollow noreferrer">ground</a> arbitrarily. It's just a reference point. I'm going to put it at the bottom of the schematic.</p> <p>With 2 amps through R1+R2, V(A) will be:</p> <p>$$V_A = R_2 \cdot 2 \ \text{A} = 4000 \ \text{V}$$</p> <p>And with 1 amp through R3+R4, V(B) will be:</p> <p>$$V_B = R_4 \cdot 1 \ \text{A} = 3000 \ \text{V}$$</p> <p>The voltage difference will be:</p> <p>$$v_o = V_A - V_B = 1000 \ \text{V}$$</p> <p>This is easily confirmed by a quick simulation:</p> <p><img src="https://i.stack.imgur.com/BLTT3.png" alt="schematic"></p> <p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fBLTT3.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p> <p>Click to open the schematic, then run the DC simulation. You'll see the <a href="https://www.circuitlab.com/docs/expressions/" rel="nofollow noreferrer">expression</a> "V(A) - V(B)" computed as 1.000 kV.</p>