The circuit in the figure below has been connected for a...

The circuit in the figure below has been connected for a long time. Let R1 = 8.40 ? and R2 = 5.60 ?. 10.0 V 1.00 ? 1.00 ?F R1 R2 2.00 ? (a) What is the potential difference across the capacitor? 6.56 V (b) If the battery is disconnected from the circuit, over what time interval does the capacitor discharge to one-half its initial voltage? 1.5 ?s Your response differs from the correct answer by more than 10%. Double check your calculations.

Transcript

00:01 In this question, we have to find the potential difference across the capacitor in the given circuit.

00:06 First what we will assume is that this is point va and this is point vb.

00:12 Now let's simplify the circuit.

00:15 The current will pass through here.

00:17 We'll split here in i1 and i2.

00:21 We can see that the current i1 is passing through 1 om and 5 .6 oom.

00:28 So these are in cd.

00:30 So we will write r series for 1 plus is equal to 1 plus 5 .60 which is 6 .60.

00:40 Similarly series in second here 8 .40 and 2 oom are in series 8 .40 plus 2 which is 10 .40.

00:51 Let's redraw the circuit.

00:54 We will assume that there is no capacitor for now.

00:58 This is 6 .60 and this is 10 .40.

01:03 Now these are in parallel.

01:04 So we will simplify the circuit.

01:07 The parallel equivalent resistance will come out to be r1 multiplied by r2 divided by r1 plus r2 which is equal to 6 .60 multiplied by 10 .40 divided by 6 .60 plus 10 .40.

01:25 The equivalent parallel resistance will be equal to 4 ome, approximately equal to 4 ome.

01:34 Now that we have found the equivalent resistance, we will find the current traveling towards the circuit.

01:42 Let's say i.

01:43 So v is equal to i r.

01:46 The voltage of battery is 10 volt and i we have to find out the equivalent resistance is equal to 4 om.

01:53 Current passing through the circuit will be 10 divided by a.

01:56 Which is 2 .5 oms.

01:58 The current in the circuit will be 2 .5 oms.

02:02 Now at point x the current is divided into i1 and i2.

02:08 We will calculate i1 and i2 respectively.

02:12 I1 will be equal to i multiplied by the resistance rs1, sorry, rs2 divided by rs1 plus rs2.

02:26 I1 will be equal to 2 .5 multiplied by 10 .40 divided by 10 .40 plus 6 .60...

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