For the circuit shown in Fig. 30–35, show that if the conditio...

Question

For the circuit shown in Fig. 30–35, show that if the condition R₁ R₂ = L/C is satisfied then the potential difference between points a and b is zero for all frequencies.

Accepted Answer

Welcome back. Everyone in this problem for the circuit below determine the relationship between circuit elements that makes the potential difference between the points M and N zero. Start by analyzing the potential difference between these points along two different paths. For answer choices. A says R divided by RB equals the square root of LCB says RARB equals the square root of L divided by CC says R divided by RB equals LC. And the D says Rab equals L divided by C. Now, in this problem, if we are going to determine the relationship between our elements for a particular condition for a potential difference, that is for the potential difference between the points M and N to be zero, then we'll have to start by figuring out what currents are going through both of our branches. So let's go ahead and do that. OK? No, to start. OK. That means we know that our R MS current through branch A. So let's call that A R MS is going to be equal to our RMS voltage divided by the impedance, which in this case is going to be the square root of R squared plus our in our capacity reactants XC squared. Likewise, for our branch with our resistor RB, that is our middle branch. OK. Then the R MS current through this branch is going to be equal to the R MS voltage divided by the square root of RB squared plus or inductive reactants squared. OK. So now we have the currents for both of our branches. Since we have that, then we can start to think about the potential difference. But again, let's think about our condition. We know that we want, we want the relationship such that the potential difference between the points M and N is zero. And if we look at our circuit, there are two ways to get from M to N, we could go from our capacitor and then through our resistor RB or we could go through our resistor R A and then through our indoor L. So let's set both of these paths equal to zero for the potential difference and try to get some expressions that will help us to solve. Let's start with the first path. OK. Let me put that in red here. That is for the path from RB to C. Now, for that path, remember we want to find it such that the potential difference equals zero. So this potential difference which is going to be I A MS multiplied by XC minus IB RMS multiplied by RB should be equal to zero. Now, if we substitute our values that we know for our current. Uh That means it's going to be equal to VR MS divided by the square root of R A plus XC squared multiplied by XC minus VR MS divided by the square root. OK. Of RB squared plus XL fix that here XL squared multiplied by RB is equal to zero. Now, we could equate both of these and cancel all our expressions for our R MS voltage. OK? Because now that we've set it to zero by canceling and rearing. Then we should find that we get XC multiplied by the square root of RB squared plus XL squared. OK is equal to RB multiplied by the square root of R squared plus XC squared. Let's call this equation one. Now that we have this equation, let's move on to our second part that is for R A to L OK? Now let me put that in blue here. Now what's the potential difference going to be? But again, we know it must be equal to zero. So now this tells us then that I be mhm multiplied by XL minus I a multiplied by X A should be equal to zero. Let's substitute our values for the current. And again, try to get an expression or try to get an equation like equation one. No, that means then that this is going to be the R MS divided by the square root of RB squared plus XL squared multiplied by XL minus KVR MS divided by the square root of R A squared plus XC squared multiplied by R A equals zero. Now, we can cancel our R MS voltage from both terms. And let me carry it over here. When we rearrange similar to equation one, then we should find that we get R multiplied by the square root of RB squared plus XL squared. OK. Equal to XL multiplied by the square root of R squared plus XC squared. Now again, we can call this equation two. Now know that we have both of these equations. How can we simplify it to figure out the relationship between the elements? Well, we can divide equation one by equation two. OK? And now let me put this in in black here when we divide dividing equation one by two, OK. If we think about it here on our left hand side, we're going well, you know what, let me, let me take, let me copy both of these equations just so that we can see exactly what's going on. So if we divide equation one by equation two, we know this is equation one. OK? And this is equation two that I'm gonna put right here, okay? And let me just clear up some of the stuff here, right? So that's XC. And let's just try to make it look a bit neater. Let me carry this over here. OK? And now when we divide, let me put that the division line in black here. When we divide equation one by equation two, what do you notice? Well, the square root of RB squared plus XL squared cancels. OK. Actually, this should be separate here. So both of these cancel. And likewise, on the right hand side, both of our root terms cancel. What does that leave us with? Well, no, this tells us then that XC divided by R A is equal to RB divided by XL. OK. When we rearrange, we get this as XC XL being equal to Rarb. But now we're not finished yet. What do we know about XC? And XL? Well, we know that XC is our capacity reactant one divided by Omega C and we know XL is our inductive reactance which is equal to Omega L. Then that tells us then that Rab is going to be equal to Omega L divided by Omega C which means then that if we cancel Omega, then rab equals L divided by C. Now this, this result, this result gives us the required relationship to achieve the poten to make the potential difference between the points M and N equals zero. In other words, the product of our resistances must be equal to the ratio of the inductance to the capacitance. If we look back at our answer choices that tells us, then that D is the correct answer. Thanks a lot for watching everyone. I hope this video helped

For the circuit shown in Fig. 30–35, show that if the condition R₁ R₂ = L/C is satisfied then the potential difference between points a and b is zero for all frequencies.

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Key Concepts

Resonance in RLC Circuits

Voltage Division in Circuits

Impedance in AC Circuits

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