Electrical Lamp Based MCQs | Brighter Darker Bulb | Series Parallel | SSCJE UPPCL MSEDCL |हिंदी 🔴

[@Suraj-xm7ws]

Thank you so much dear Sir Aapne sir doubt clear kr diye🙏

[@rajatsharma9806]

Really Nice. 100% Recommendable.

[@jayshreeram7060]

Very nice class sir 👌

[@purajitsarkar9861]

Answer will be B.60 Watt Exp. P=(100×150/250);=60 W

[@ramguin4932]

please explain this equation briefly ....

[@Self_study_with_me]

60w right ans.☑️☑️

[@sakshigupta4793]

Option b 60W

[@SathishKumar-br9xn]

Nice questions sir cleared my doubts, thank you.

[@gaurav1647]

Ans. 60 W

[@aktar9609]

Really helpful sir ❤️ ❤️ ❤️

[@SathishKumar-br9xn]

And: 60W

[@yashpalshukla5662]

60W

[@rupeshwaridiwan2843]

Thank you sir🙏

[@ElectricalEngineeringMCQs]

Most welcome, Kindly DO Share this session with your friends.

[@ayushikumari9130]

60 W

[@puneetsharma7470]

Ach h

[@shakeelahmad9370]

Sir please Compare 2 3 number question, in second question bulb B have higher resistance and glow brighter but u said in 3rd question which bulb have have higher resistance will glow dim....please clear this things

[@souravdey9278]

The 2nd question says the bulbs are in series and the 3rd question our household wirings are done in parallel . So, the power dissipation is inversely proportional to the resistance. In series the higher resistance bulb will be brighter In parallel the lesser resistance bulb will be brighter

[@dipanshrajput9021]

60 w

[@RAHULTIWARI-gc8hx]

60 Watt, sir.

[@ElectricalEngineeringMCQs]

Kindly DO Share this session with your friends.

[@narenderreddy7305]

Okay sir

[@Saurabhkumar-gq2vd]

Very very helpful video🙏🙏...Sir electric traction and engiring materials par bhi video banao please.. sir

[@ElectricalEngineeringMCQs]

Sure Saurabh I will try to make it soon possible Kindly Stay tuned with us and DO Share this session with your friends.

[@Saurabhkumar-gq2vd]

Ok sir

[@lucykumari5703]

Sir 6 no ques me agar voltage half naii rahta...or different rahta to...total power nikalne k lie (n)² se multiply krte kya....kyuki 4 no ques me to (n)² se multiply naii kie....???

[@SatishKumar-cj4wo]

Kyuki usme voltage barabar tha, difference ratio 1 tha , yani n=1

[@lucykumari5703]

Oo thanks but agar dono lamp ka voltage different rahta 6 no me to kya krte...

[@ElectricalEngineeringMCQs]

I will consider this question in my upcoming session.

[@theart9371]

Sir apke first question me two bulb are in series and a 300 volt ka voltage source connected hai to dono bulb ka voltage 300 ho skta hai?

[@SurajKumar-qv6jf]

P=p1*p2/p1+p2=60w

[@mohitchauhan1185]

Thank-you♥

[@ElectricalEngineeringMCQs]

Most welcome Mohit, Kindly DO Share this session with your friends.

[@er.vermajaipur6604]

60watt

[@HashimKhan-hk1zj]

60 watt

[@knowledge21002]

Ans. 60 watt

[@S.j-ze3nz]

Current.exam k.q.s prvide sir.

[@rajendrashahrj4256]

Wow appreciate you hard work 👍🙏🙏🙏❤️💯

[@ElectricalEngineeringMCQs]

I appreciate that Rajendra, Kindly DO Share this session with your friends.

[@lokeshprajapati-2668]

Sir electric welding par bhi MCQ provide karwa dijiye please

[@ElectricalEngineeringMCQs]

Sure.

[@fitnessworld2493]

60w

[@533_manjitmali9]

👍👍❤

[@RahulKumar-lq8zp]

Sir, R = V/I hota hai, aap v square upon I kyu lai rahe ho?

[@srikarjan5778]

Sir..in 2nd question summary it is said that if i2R losses is more...the bulb will glow brighter.....then in 3rd question the brighter bulb should be having more resistance then the dim bulb na sir.......plz correct me if I’m wrong........thank you

[@ElectricalEngineeringMCQs]

Very Interesting Fact Srikar, I appreciate your findings See, here you might have noticed that in 2nd Q. the connection is series and thus the same current flow through the bulbs and here comes the difference in Q.3 where we have parallel connection where you should know that current will always try to find the least resistance path. I think this might clear your doubt.

[@srikarjan5778]

@@ElectricalEngineeringMCQs ok sir doubt is clarified....thank you for the early response.

[@ElectricalEngineeringMCQs]

Indeed, I appreciate In return you would share this session with your friends as well. Stay tuned and Stay safe.

[@mansoorqazi3582]

In this question sir did mistake.

[@rndefance]

60watt.

[@ravindrachouhan7836]

Sir pdf provide kijiye

[@prembhukya5785]

Hamare bulb 💡 ko bhati jaladiya sirji

[@ElectricalEngineeringMCQs]

kindly DO Share this session with your friends.

[@babusatavekar4977]

Hello sir, I am the your subscriber. I watches your every videos. You are uploading very helpful videos. Thanks for this. And you are my final solution , if i confused to any of the question. But now, I have a doubt Q. If there are two lamps are connented in series given with wattage and supply source then how we can find the resistance, voltage drop and current. There are only voltage and wattage is given. Pls reply me sir. I am waiting for your reply.

[@ElectricalEngineeringMCQs]

I'm so glad to know Babu Satavekar that you are consistently watching all my session indeed they are useful as well. Keep continue watching them and DO Share with your friends. Now, Replying to question See If the Power and Voltage is given then 1. you do calculate the resistance as I have explained in the session, 2. DO find out the total resistance i,e if they are in series they gets added. 3. calculate the total current and multiply it with the respective resistances to calculate the voltage drop. I hope you understood this steps, if not I will surely try to cover this question in my next session. Kindly Stay tuned and Stay safe.

[@babusatavekar4977]

@@ElectricalEngineeringMCQs Sir, But, there is not given the values of resistance of lamp and current taken by lamp. So how can we find it. As you explained in this session if we go to find the value of resistance then we have to need the value of voltage and current (As formula V^2/P). Or if we go to find the value of current the we have to need the value of voltage and resistance. Without them we can't do it. As circuit law 1. In series circuit voltage distributed. 2. In parallel circuit current distributed.

[@ElectricalEngineeringMCQs]

I think I might have explained the same thing in Question no 7 at 21:01, I have taken up your feedback and will try to explain the same in my next session. Moreover, Even if we have the power and voltage ratings we can calculate the resistance and so the rated current (I=V/R). There has to be at least two values or some kind of interdependency which might change according to the question to question. Yet I have found your query very appreciable. I hope you must have shared this session with your friends. Stay tuned.

[@Rohitkumar-mp3fv]

In this condition every bulb's rated voltage is considered equal to source voltage. I think yhi puchhna cha rhe thhe aap

[@puneetsharma7470]

Video

[@barconessgaming9826]

100*150/250 =60W

[@Rohit_ahire97]

Q 7 not good explain Try improve explain briefly

[@amitbhatia1998]

Thanks sir❤ Good explain...🙏

[@mandarisasare8390]

60 W

[@me-ps-0693]

60W

[@sandeepyadavofficial523]

60 w

[@electro2138]

60 watt

[@mounikaadepu2150]

60W

[@SUBHAJITCHAKRABORTY-me3sr]

60 W

[@arunkm3438]

60W

[@electricpanelmake]

60 watt

[@virenderthakur9791]

66 watt

[@vinaykumaryadav6143]

60W

[@nidhirajyaguru6039]

60W